Calculus Examples

$lim_{x \to \frac{1}{2}} \frac{(x - \frac{1}{2}) (6 x^{2} + x - 2)}{4 x^{2} - 4 x + 1}$

Step 1

Combine terms.

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Step 1.1

To write $x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{(x \cdot \frac{2}{2} - \frac{1}{2}) (6 x^{2} + x - 2)}{4 x^{2} - 4 x + 1}$

Step 1.2

Combine $x$ and $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{(\frac{x \cdot 2}{2} - \frac{1}{2}) (6 x^{2} + x - 2)}{4 x^{2} - 4 x + 1}$

Step 1.3

Combine the numerators over the common denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{x \cdot 2 - 1}{2} (6 x^{2} + x - 2)}{4 x^{2} - 4 x + 1}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 2.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to \frac{1}{2}} \frac{x \cdot 2 - 1}{2} (6 x^{2} + x - 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{lim_{x \to \frac{1}{2}} \frac{x \cdot 2 - 1}{2} \cdot lim_{x \to \frac{1}{2}} 6 x^{2} + x - 2}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{\frac{1}{2} lim_{x \to \frac{1}{2}} x \cdot 2 - 1 \cdot lim_{x \to \frac{1}{2}} 6 x^{2} + x - 2}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{\frac{1}{2} (lim_{x \to \frac{1}{2}} x \cdot 2 - lim_{x \to \frac{1}{2}} 1) \cdot lim_{x \to \frac{1}{2}} 6 x^{2} + x - 2}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 1) \cdot lim_{x \to \frac{1}{2}} 6 x^{2} + x - 2}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.5

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1) \cdot lim_{x \to \frac{1}{2}} 6 x^{2} + x - 2}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1) \cdot (lim_{x \to \frac{1}{2}} 6 x^{2} + lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.7

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1) \cdot (6 lim_{x \to \frac{1}{2}} x^{2} + lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.8

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1) \cdot (6 {(lim_{x \to \frac{1}{2}} x)}^{2} + lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.9

Evaluate the limit of $2$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{\frac{1}{2} (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1) \cdot (6 {(lim_{x \to \frac{1}{2}} x)}^{2} + lim_{x \to \frac{1}{2}} x - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.10

Evaluate the limits by plugging in $\frac{1}{2}$ for all occurrences of $x$ .

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Step 2.1.2.10.1

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{\frac{1}{2} (2 (\frac{1}{2}) - 1 \cdot 1) \cdot (6 {(lim_{x \to \frac{1}{2}} x)}^{2} + lim_{x \to \frac{1}{2}} x - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.10.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{\frac{1}{2} (2 (\frac{1}{2}) - 1 \cdot 1) \cdot (6 {(\frac{1}{2})}^{2} + lim_{x \to \frac{1}{2}} x - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.10.3

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{\frac{1}{2} (2 (\frac{1}{2}) - 1 \cdot 1) \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11

Simplify the answer.

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Step 2.1.2.11.1

Simplify each term.

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Step 2.1.2.11.1.1

Cancel the common factor of $2$ .

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Step 2.1.2.11.1.1.1

Cancel the common factor.

$\frac{\frac{1}{2} (2 (\frac{1}{2}) - 1 \cdot 1) \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.1.1.2

Rewrite the expression.

$\frac{\frac{1}{2} (1 - 1 \cdot 1) \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.1.2

Multiply $- 1$ by $1$ .

$\frac{\frac{1}{2} (1 - 1) \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.2

Subtract $1$ from $1$ .

$\frac{\frac{1}{2} \cdot 0 \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.3

Multiply $\frac{1}{2}$ by $0$ .

$\frac{0 \cdot (6 {(\frac{1}{2})}^{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4

Simplify each term.

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Step 2.1.2.11.4.1

Apply the product rule to $\frac{1}{2}$ .

$\frac{0 \cdot (6 \frac{1^{2}}{2^{2}} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.2

One to any power is one.

$\frac{0 \cdot (6 \frac{1}{2^{2}} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.3

Raise $2$ to the power of $2$ .

$\frac{0 \cdot (6 (\frac{1}{4}) + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.4

Cancel the common factor of $2$ .

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Step 2.1.2.11.4.4.1

Factor $2$ out of $6$ .

$\frac{0 \cdot (2 (3) \frac{1}{4} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.4.2

Factor $2$ out of $4$ .

$\frac{0 \cdot (2 \cdot 3 \frac{1}{2 \cdot 2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.4.3

Cancel the common factor.

$\frac{0 \cdot (2 \cdot 3 \frac{1}{2 \cdot 2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.4.4

Rewrite the expression.

$\frac{0 \cdot (3 (\frac{1}{2}) + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.5

Combine $3$ and $\frac{1}{2}$ .

$\frac{0 \cdot (\frac{3}{2} + \frac{1}{2} - 1 \cdot 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.4.6

Multiply $- 1$ by $2$ .

$\frac{0 \cdot (\frac{3}{2} + \frac{1}{2} - 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.5

Combine the numerators over the common denominator.

$\frac{0 \cdot (- 2 + \frac{3 + 1}{2})}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.6

Add $3$ and $1$ .

$\frac{0 \cdot (- 2 + \frac{4}{2})}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.7

Divide $4$ by $2$ .

$\frac{0 \cdot (- 2 + 2)}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.8

Add $- 2$ and $2$ .

$\frac{0 \cdot 0}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.2.11.9

Multiply $0$ by $0$ .

$\frac{0}{lim_{x \to \frac{1}{2}} 4 x^{2} - 4 x + 1}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{0}{lim_{x \to \frac{1}{2}} 4 x^{2} - lim_{x \to \frac{1}{2}} 4 x + lim_{x \to \frac{1}{2}} 1}$

Step 2.1.3.2

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{4 lim_{x \to \frac{1}{2}} x^{2} - lim_{x \to \frac{1}{2}} 4 x + lim_{x \to \frac{1}{2}} 1}$

Step 2.1.3.3

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{0}{4 {(lim_{x \to \frac{1}{2}} x)}^{2} - lim_{x \to \frac{1}{2}} 4 x + lim_{x \to \frac{1}{2}} 1}$

Step 2.1.3.4

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{4 {(lim_{x \to \frac{1}{2}} x)}^{2} - 4 lim_{x \to \frac{1}{2}} x + lim_{x \to \frac{1}{2}} 1}$

Step 2.1.3.5

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{0}{4 {(lim_{x \to \frac{1}{2}} x)}^{2} - 4 lim_{x \to \frac{1}{2}} x + 1}$

Step 2.1.3.6

Evaluate the limits by plugging in $\frac{1}{2}$ for all occurrences of $x$ .

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Step 2.1.3.6.1

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{0}{4 {(\frac{1}{2})}^{2} - 4 lim_{x \to \frac{1}{2}} x + 1}$

Step 2.1.3.6.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{0}{4 {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7

Simplify the answer.

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Step 2.1.3.7.1

Simplify each term.

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Step 2.1.3.7.1.1

Apply the product rule to $\frac{1}{2}$ .

$\frac{0}{4 \frac{1^{2}}{2^{2}} - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7.1.2

One to any power is one.

$\frac{0}{4 \frac{1}{2^{2}} - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7.1.3

Raise $2$ to the power of $2$ .

$\frac{0}{4 (\frac{1}{4}) - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7.1.4

Cancel the common factor of $4$ .

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Step 2.1.3.7.1.4.1

Cancel the common factor.

$\frac{0}{4 (\frac{1}{4}) - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7.1.4.2

Rewrite the expression.

$\frac{0}{1 - 4 (\frac{1}{2}) + 1}$

Step 2.1.3.7.1.5

Cancel the common factor of $2$ .

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Step 2.1.3.7.1.5.1

Factor $2$ out of $- 4$ .

$\frac{0}{1 + 2 (- 2) \frac{1}{2} + 1}$

Step 2.1.3.7.1.5.2

Cancel the common factor.

$\frac{0}{1 + 2 \cdot - 2 \frac{1}{2} + 1}$

Step 2.1.3.7.1.5.3

Rewrite the expression.

$\frac{0}{1 - 2 + 1}$

Step 2.1.3.7.2

Subtract $2$ from $1$ .

$\frac{0}{- 1 + 1}$

Step 2.1.3.7.3

Add $- 1$ and $1$ .

$\frac{0}{0}$

Step 2.1.3.7.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.1.3.8

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to \frac{1}{2}} \frac{\frac{x \cdot 2 - 1}{2} (6 x^{2} + x - 2)}{4 x^{2} - 4 x + 1} = lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [\frac{x \cdot 2 - 1}{2} (6 x^{2} + x - 2)]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [\frac{x \cdot 2 - 1}{2} (6 x^{2} + x - 2)]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.2

Move $2$ to the left of $x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [\frac{2 x - 1}{2} (6 x^{2} + x - 2)]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \frac{2 x - 1}{2}$ and $g (x) = 6 x^{2} + x - 2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} \frac{d}{d x} [6 x^{2} + x - 2] + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.4

By the Sum Rule, the derivative of $6 x^{2} + x - 2$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.5

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (6 (2 x) + \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.7

Multiply $2$ by $6$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1 + \frac{d}{d x} [- 2]) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.9

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1 + 0) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.10

Add $12 x + 1$ and $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{d}{d x} [\frac{2 x - 1}{2}]}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.11

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{2 x - 1}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [2 x - 1]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) (\frac{1}{2} \frac{d}{d x} [2 x - 1])}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.12

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1])}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.13

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1])}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.14

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} (2 \cdot 1 + \frac{d}{d x} [- 1])}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.15

Multiply $2$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} (2 + \frac{d}{d x} [- 1])}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.16

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} (2 + 0)}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.17

Add $2$ and $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{1}{2} \cdot 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.18

Combine $2$ and $\frac{1}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{2}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.19

Cancel the common factor of $2$ .

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Step 2.3.19.1

Cancel the common factor.

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \frac{2}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.19.2

Rewrite the expression.

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + (6 x^{2} + x - 2) \cdot 1}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.20

Multiply $6 x^{2} + x - 2$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x + 1) + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21

Simplify.

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Step 2.3.21.1

Apply the distributive property.

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x - 1}{2} (12 x) + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2

Combine terms.

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Step 2.3.21.2.1

Combine $12$ and $\frac{2 x - 1}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1)}{2} x + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.2

Combine $\frac{12 (2 x - 1)}{2}$ and $x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x}{2} + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.3

Cancel the common factor of $12$ and $2$ .

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Step 2.3.21.2.3.1

Factor $2$ out of $12 (2 x - 1) x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 (6 (2 x - 1) x)}{2} + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.3.2

Cancel the common factors.

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Step 2.3.21.2.3.2.1

Factor $2$ out of $2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{2 (6 (2 x - 1) x)}{2 (1)} + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.3.2.2

Cancel the common factor.

$lim_{x \to \frac{1}{2}} \frac{\frac{2 (6 (2 x - 1) x)}{2 \cdot 1} + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.3.2.3

Rewrite the expression.

$lim_{x \to \frac{1}{2}} \frac{\frac{6 (2 x - 1) x}{1} + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.3.2.4

Divide $6 (2 x - 1) x$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{6 (2 x - 1) x + \frac{2 x - 1}{2} \cdot 1 + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.4

Multiply $\frac{2 x - 1}{2}$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{6 (2 x - 1) x + \frac{2 x - 1}{2} + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.5

To write $6 (2 x - 1) x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{6 (2 x - 1) x \cdot \frac{2}{2} + \frac{2 x - 1}{2} + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.6

Combine $6 (2 x - 1) x$ and $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{6 (2 x - 1) x \cdot 2}{2} + \frac{2 x - 1}{2} + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.7

Combine the numerators over the common denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{6 (2 x - 1) x \cdot 2 + 2 x - 1}{2} + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.8

Multiply $2$ by $6$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1}{2} + 6 x^{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.9

To write $6 x^{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1}{2} + 6 x^{2} \cdot \frac{2}{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.10

Combine $6 x^{2}$ and $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1}{2} + \frac{6 x^{2} \cdot 2}{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.11

Combine the numerators over the common denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 6 x^{2} \cdot 2}{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.12

Multiply $2$ by $6$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 12 x^{2}}{2} + x - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.13

To write $x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 12 x^{2}}{2} + x \cdot \frac{2}{2} - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.14

Combine $x$ and $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 12 x^{2}}{2} + \frac{x \cdot 2}{2} - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.15

Combine the numerators over the common denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 12 x^{2} + x \cdot 2}{2} - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.16

Move $2$ to the left of $x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 2 x - 1 + 12 x^{2} + 2 \cdot x}{2} - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.17

Add $2 x$ and $2 x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x - 1 + 12 x^{2}}{2} - 2}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.18

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x - 1 + 12 x^{2}}{2} - 2 \cdot \frac{2}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.19

Combine $- 2$ and $\frac{2}{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x - 1 + 12 x^{2}}{2} + \frac{- 2 \cdot 2}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.20

Combine the numerators over the common denominator.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x - 1 + 12 x^{2} - 2 \cdot 2}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.21

Multiply $- 2$ by $2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x - 1 + 12 x^{2} - 4}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.2.22

Subtract $4$ from $- 1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 (2 x - 1) x + 4 x + 12 x^{2} - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.3

Reorder terms.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 x (2 x - 1) + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4

Simplify the numerator.

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Step 2.3.21.4.1

Apply the distributive property.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 x (2 x) + 12 x \cdot - 1 + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.2

Rewrite using the commutative property of multiplication.

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 \cdot 2 x \cdot x + 12 x \cdot - 1 + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.3

Multiply $- 1$ by $12$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 \cdot 2 x \cdot x - 12 x + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.4

Simplify each term.

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Step 2.3.21.4.4.1

Multiply $x$ by $x$ by adding the exponents.

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Step 2.3.21.4.4.1.1

Move $x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 \cdot 2 (x \cdot x) - 12 x + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.4.1.2

Multiply $x$ by $x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 12 \cdot 2 x^{2} - 12 x + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.4.2

Multiply $12$ by $2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{12 x^{2} + 24 x^{2} - 12 x + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.5

Add $12 x^{2}$ and $24 x^{2}$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{36 x^{2} - 12 x + 4 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.6

Add $- 12 x$ and $4 x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{36 x^{2} - 8 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7

Factor by grouping.

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Step 2.3.21.4.7.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 36 \cdot - 5 = - 180$ and whose sum is $b = - 8$ .

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Step 2.3.21.4.7.1.1

Factor $- 8$ out of $- 8 x$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{36 x^{2} - 8 (x) - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7.1.2

Rewrite $- 8$ as $10$ plus $- 18$

$lim_{x \to \frac{1}{2}} \frac{\frac{36 x^{2} + (10 - 18) x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7.1.3

Apply the distributive property.

$lim_{x \to \frac{1}{2}} \frac{\frac{36 x^{2} + 10 x - 18 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7.2

Factor out the greatest common factor from each group.

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Step 2.3.21.4.7.2.1

Group the first two terms and the last two terms.

$lim_{x \to \frac{1}{2}} \frac{\frac{(36 x^{2} + 10 x) - 18 x - 5}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7.2.2

Factor out the greatest common factor (GCF) from each group.

$lim_{x \to \frac{1}{2}} \frac{\frac{2 x (18 x + 5) - (18 x + 5)}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.21.4.7.3

Factor the polynomial by factoring out the greatest common factor, $18 x + 5$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{\frac{d}{d x} [4 x^{2} - 4 x + 1]}$

Step 2.3.22

By the Sum Rule, the derivative of $4 x^{2} - 4 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]}$

Step 2.3.23

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Step 2.3.23.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]}$

Step 2.3.23.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{4 (2 x) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]}$

Step 2.3.23.3

Multiply $2$ by $4$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [1]}$

Step 2.3.24

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 2.3.24.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [1]}$

Step 2.3.24.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x - 4 \cdot 1 + \frac{d}{d x} [1]}$

Step 2.3.24.3

Multiply $- 4$ by $1$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x - 4 + \frac{d}{d x} [1]}$

Step 2.3.25

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x - 4 + 0}$

Step 2.3.26

Add $8 x - 4$ and $0$ .

$lim_{x \to \frac{1}{2}} \frac{\frac{(18 x + 5) (2 x - 1)}{2}}{8 x - 4}$

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 x - 1)}{2} \cdot \frac{1}{8 x - 4}$

Step 2.5

Multiply $\frac{(18 x + 5) (2 x - 1)}{2}$ by $\frac{1}{8 x - 4}$ .

$lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 x - 1)}{2 (8 x - 4)}$

Step 3

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 x - 1)}{8 x - 4}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 4.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1}{2} \cdot \frac{lim_{x \to \frac{1}{2}} (18 x + 5) (2 x - 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{lim_{x \to \frac{1}{2}} 18 x + 5 \cdot lim_{x \to \frac{1}{2}} 2 x - 1}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{(lim_{x \to \frac{1}{2}} 18 x + lim_{x \to \frac{1}{2}} 5) \cdot lim_{x \to \frac{1}{2}} 2 x - 1}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.3

Move the term $18$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot \frac{(18 lim_{x \to \frac{1}{2}} x + lim_{x \to \frac{1}{2}} 5) \cdot lim_{x \to \frac{1}{2}} 2 x - 1}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.4

Evaluate the limit of $5$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{(18 lim_{x \to \frac{1}{2}} x + 5) \cdot lim_{x \to \frac{1}{2}} 2 x - 1}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{(18 lim_{x \to \frac{1}{2}} x + 5) \cdot (lim_{x \to \frac{1}{2}} 2 x - lim_{x \to \frac{1}{2}} 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.6

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot \frac{(18 lim_{x \to \frac{1}{2}} x + 5) \cdot (2 lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.7

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{(18 lim_{x \to \frac{1}{2}} x + 5) \cdot (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.8

Evaluate the limits by plugging in $\frac{1}{2}$ for all occurrences of $x$ .

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Step 4.1.2.8.1

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{1}{2} \cdot \frac{(18 (\frac{1}{2}) + 5) \cdot (2 lim_{x \to \frac{1}{2}} x - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.8.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{1}{2} \cdot \frac{(18 (\frac{1}{2}) + 5) \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9

Simplify the answer.

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Step 4.1.2.9.1

Cancel the common factor of $2$ .

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Step 4.1.2.9.1.1

Factor $2$ out of $18$ .

$\frac{1}{2} \cdot \frac{(2 (9) \frac{1}{2} + 5) \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.1.2

Cancel the common factor.

$\frac{1}{2} \cdot \frac{(2 \cdot 9 \frac{1}{2} + 5) \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.1.3

Rewrite the expression.

$\frac{1}{2} \cdot \frac{(9 + 5) \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.2

Add $9$ and $5$ .

$\frac{1}{2} \cdot \frac{14 \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.3

Simplify each term.

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Step 4.1.2.9.3.1

Cancel the common factor of $2$ .

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Step 4.1.2.9.3.1.1

Cancel the common factor.

$\frac{1}{2} \cdot \frac{14 \cdot (2 (\frac{1}{2}) - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.3.1.2

Rewrite the expression.

$\frac{1}{2} \cdot \frac{14 \cdot (1 - 1 \cdot 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.3.2

Multiply $- 1$ by $1$ .

$\frac{1}{2} \cdot \frac{14 \cdot (1 - 1)}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.4

Subtract $1$ from $1$ .

$\frac{1}{2} \cdot \frac{14 \cdot 0}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.2.9.5

Multiply $14$ by $0$ .

$\frac{1}{2} \cdot \frac{0}{lim_{x \to \frac{1}{2}} 8 x - 4}$

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

Evaluate the limit.

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Step 4.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{0}{lim_{x \to \frac{1}{2}} 8 x - lim_{x \to \frac{1}{2}} 4}$

Step 4.1.3.1.2

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot \frac{0}{8 lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 4}$

Step 4.1.3.1.3

Evaluate the limit of $4$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} \cdot \frac{0}{8 lim_{x \to \frac{1}{2}} x - 1 \cdot 4}$

Step 4.1.3.2

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{1}{2} \cdot \frac{0}{8 (\frac{1}{2}) - 1 \cdot 4}$

Step 4.1.3.3

Simplify the answer.

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Step 4.1.3.3.1

Simplify each term.

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Step 4.1.3.3.1.1

Cancel the common factor of $2$ .

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Step 4.1.3.3.1.1.1

Factor $2$ out of $8$ .

$\frac{1}{2} \cdot \frac{0}{2 (4) \frac{1}{2} - 1 \cdot 4}$

Step 4.1.3.3.1.1.2

Cancel the common factor.

$\frac{1}{2} \cdot \frac{0}{2 \cdot 4 \frac{1}{2} - 1 \cdot 4}$

Step 4.1.3.3.1.1.3

Rewrite the expression.

$\frac{1}{2} \cdot \frac{0}{4 - 1 \cdot 4}$

Step 4.1.3.3.1.2

Multiply $- 1$ by $4$ .

$\frac{1}{2} \cdot \frac{0}{4 - 4}$

Step 4.1.3.3.2

Subtract $4$ from $4$ .

$\frac{1}{2} \cdot \frac{0}{0}$

Step 4.1.3.3.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

Step 4.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

Step 4.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

Step 4.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 x - 1)}{8 x - 4} = lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [(18 x + 5) (2 x - 1)]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{\frac{d}{d x} [(18 x + 5) (2 x - 1)]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 18 x + 5$ and $g (x) = 2 x - 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) \frac{d}{d x} [2 x - 1] + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.3

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 \cdot 1 + \frac{d}{d x} [- 1]) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.6

Multiply $2$ by $1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 + \frac{d}{d x} [- 1]) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.7

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) (2 + 0) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.8

Add $2$ and $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{(18 x + 5) \cdot 2 + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.9

Move $2$ to the left of $18 x + 5$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 \cdot (18 x + 5) + (2 x - 1) \frac{d}{d x} [18 x + 5]}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.10

By the Sum Rule, the derivative of $18 x + 5$ with respect to $x$ is $\frac{d}{d x} [18 x] + \frac{d}{d x} [5]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) (\frac{d}{d x} [18 x] + \frac{d}{d x} [5])}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.11

Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) (18 \frac{d}{d x} [x] + \frac{d}{d x} [5])}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.12

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) (18 \cdot 1 + \frac{d}{d x} [5])}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.13

Multiply $18$ by $1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) (18 + \frac{d}{d x} [5])}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.14

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) (18 + 0)}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.15

Add $18$ and $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + (2 x - 1) \cdot 18}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.16

Move $18$ to the left of $2 x - 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 (18 x + 5) + 18 \cdot (2 x - 1)}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17

Simplify.

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Step 4.3.17.1

Apply the distributive property.

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 \cdot 18 x + 2 \cdot 5 + 18 (2 x - 1)}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.2

Apply the distributive property.

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{2 \cdot 18 x + 2 \cdot 5 + 18 \cdot 2 x + 18 \cdot - 1}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3

Combine terms.

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Step 4.3.17.3.1

Multiply $18$ by $2$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{36 x + 2 \cdot 5 + 18 \cdot 2 x + 18 \cdot - 1}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3.2

Multiply $2$ by $5$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{36 x + 10 + 18 \cdot 2 x + 18 \cdot - 1}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3.3

Multiply $2$ by $18$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{36 x + 10 + 36 x + 18 \cdot - 1}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3.4

Multiply $18$ by $- 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{36 x + 10 + 36 x - 18}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3.5

Add $36 x$ and $36 x$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x + 10 - 18}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.17.3.6

Subtract $18$ from $10$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{\frac{d}{d x} [8 x - 4]}$

Step 4.3.18

By the Sum Rule, the derivative of $8 x - 4$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 4]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{\frac{d}{d x} [8 x] + \frac{d}{d x} [- 4]}$

Step 4.3.19

Evaluate $\frac{d}{d x} [8 x]$ .

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Step 4.3.19.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{8 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]}$

Step 4.3.19.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{8 \cdot 1 + \frac{d}{d x} [- 4]}$

Step 4.3.19.3

Multiply $8$ by $1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{8 + \frac{d}{d x} [- 4]}$

Step 4.3.20

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{8 + 0}$

Step 4.3.21

Add $8$ and $0$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{72 x - 8}{8}$

Step 4.4

Cancel the common factor of $72 x - 8$ and $8$ .

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Step 4.4.1

Factor $8$ out of $72 x$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{8 \cdot 9 x - 8}{8}$

Step 4.4.2

Factor $8$ out of $- 8$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{8 \cdot 9 x + 8 \cdot - 1}{8}$

Step 4.4.3

Factor $8$ out of $8 (9 x) + 8 (- 1)$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{8 (9 x - 1)}{8}$

Step 4.4.4

Cancel the common factors.

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Step 4.4.4.1

Factor $8$ out of $8$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{8 (9 x - 1)}{8 (1)}$

Step 4.4.4.2

Cancel the common factor.

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{8 (9 x - 1)}{8 \cdot 1}$

Step 4.4.4.3

Rewrite the expression.

$\frac{1}{2} lim_{x \to \frac{1}{2}} \frac{9 x - 1}{1}$

Step 4.4.4.4

Divide $9 x - 1$ by $1$ .

$\frac{1}{2} lim_{x \to \frac{1}{2}} 9 x - 1$

Step 5

Evaluate the limit.

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Step 5.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} (lim_{x \to \frac{1}{2}} 9 x - lim_{x \to \frac{1}{2}} 1)$

Step 5.2

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} (9 lim_{x \to \frac{1}{2}} x - lim_{x \to \frac{1}{2}} 1)$

Step 5.3

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{1}{2}$ .

$\frac{1}{2} (9 lim_{x \to \frac{1}{2}} x - 1 \cdot 1)$

Step 6

Evaluate the limit of $x$ by plugging in $\frac{1}{2}$ for $x$ .

$\frac{1}{2} (9 (\frac{1}{2}) - 1 \cdot 1)$

Step 7

Simplify the answer.

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Step 7.1

Simplify each term.

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Step 7.1.1

Combine $9$ and $\frac{1}{2}$ .

$\frac{1}{2} (\frac{9}{2} - 1 \cdot 1)$

Step 7.1.2

Multiply $- 1$ by $1$ .

$\frac{1}{2} (\frac{9}{2} - 1)$

Step 7.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} (\frac{9}{2} - 1 \cdot \frac{2}{2})$

Step 7.3

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} (\frac{9}{2} + \frac{- 1 \cdot 2}{2})$

Step 7.4

Combine the numerators over the common denominator.

$\frac{1}{2} \cdot \frac{9 - 1 \cdot 2}{2}$

Step 7.5

Simplify the numerator.

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Step 7.5.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} \cdot \frac{9 - 2}{2}$

Step 7.5.2

Subtract $2$ from $9$ .

$\frac{1}{2} \cdot \frac{7}{2}$

Step 7.6

Multiply $\frac{1}{2} \cdot \frac{7}{2}$ .

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Step 7.6.1

Multiply $\frac{1}{2}$ by $\frac{7}{2}$ .

$\frac{7}{2 \cdot 2}$

Step 7.6.2

Multiply $2$ by $2$ .

$\frac{7}{4}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{7}{4}$

Decimal Form:

$1.75$