Calculus Examples

$lim_{x \to 1} \frac{\log (x)}{7 x^{2} + 5 x - 12}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to 1} \log (x)}{lim_{x \to 1} 7 x^{2} + 5 x - 12}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Move the limit inside the logarithm.

$\frac{\log (lim_{x \to 1} x)}{lim_{x \to 1} 7 x^{2} + 5 x - 12}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{\log (1)}{lim_{x \to 1} 7 x^{2} + 5 x - 12}$

Step 1.2.3

Logarithm base $10$ of $1$ is $0$ .

$\frac{0}{lim_{x \to 1} 7 x^{2} + 5 x - 12}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{0}{lim_{x \to 1} 7 x^{2} + lim_{x \to 1} 5 x - lim_{x \to 1} 12}$

Step 1.3.2

Move the term $7$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{7 lim_{x \to 1} x^{2} + lim_{x \to 1} 5 x - lim_{x \to 1} 12}$

Step 1.3.3

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

$\frac{0}{7 {(lim_{x \to 1} x)}^{2} + lim_{x \to 1} 5 x - lim_{x \to 1} 12}$

Step 1.3.4

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{7 {(lim_{x \to 1} x)}^{2} + 5 lim_{x \to 1} x - lim_{x \to 1} 12}$

Step 1.3.5

Evaluate the limit of $12$ which is constant as $x$ approaches $1$ .

$\frac{0}{7 {(lim_{x \to 1} x)}^{2} + 5 lim_{x \to 1} x - 1 \cdot 12}$

Step 1.3.6

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 1.3.6.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{0}{7 \cdot 1^{2} + 5 lim_{x \to 1} x - 1 \cdot 12}$

Step 1.3.6.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{0}{7 \cdot 1^{2} + 5 \cdot 1 - 1 \cdot 12}$

Step 1.3.7

Simplify the answer.

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Step 1.3.7.1

Simplify each term.

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Step 1.3.7.1.1

One to any power is one.

$\frac{0}{7 \cdot 1 + 5 \cdot 1 - 1 \cdot 12}$

Step 1.3.7.1.2

Multiply $7$ by $1$ .

$\frac{0}{7 + 5 \cdot 1 - 1 \cdot 12}$

Step 1.3.7.1.3

Multiply $5$ by $1$ .

$\frac{0}{7 + 5 - 1 \cdot 12}$

Step 1.3.7.1.4

Multiply $- 1$ by $12$ .

$\frac{0}{7 + 5 - 12}$

Step 1.3.7.2

Add $7$ and $5$ .

$\frac{0}{12 - 12}$

Step 1.3.7.3

Subtract $12$ from $12$ .

$\frac{0}{0}$

Step 1.3.7.4

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

Undefined

$\frac{0}{0}$

Step 1.3.8

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{x \to 1} \frac{\log (x)}{7 x^{2} + 5 x - 12} = lim_{x \to 1} \frac{\frac{d}{d x} [\log (x)]}{\frac{d}{d x} [7 x^{2} + 5 x - 12]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 1} \frac{\frac{d}{d x} [\log (x)]}{\frac{d}{d x} [7 x^{2} + 5 x - 12]}$

Step 3.2

The derivative of $\log (x)$ with respect to $x$ is $\frac{1}{x \ln (10)}$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{\frac{d}{d x} [7 x^{2} + 5 x - 12]}$

Step 3.3

By the Sum Rule, the derivative of $7 x^{2} + 5 x - 12$ with respect to $x$ is $\frac{d}{d x} [7 x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 12]$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{\frac{d}{d x} [7 x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 12]}$

Step 3.4

Evaluate $\frac{d}{d x} [7 x^{2}]$ .

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Step 3.4.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x^{2}$ with respect to $x$ is $7 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{7 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 12]}$

Step 3.4.2

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

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{7 (2 x) + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 12]}$

Step 3.4.3

Multiply $2$ by $7$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 12]}$

Step 3.5

Evaluate $\frac{d}{d x} [5 x]$ .

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Step 3.5.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + 5 \frac{d}{d x} [x] + \frac{d}{d x} [- 12]}$

Step 3.5.2

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

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + 5 \cdot 1 + \frac{d}{d x} [- 12]}$

Step 3.5.3

Multiply $5$ by $1$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + 5 + \frac{d}{d x} [- 12]}$

Step 3.6

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + 5 + 0}$

Step 3.7

Add $14 x + 5$ and $0$ .

$lim_{x \to 1} \frac{\frac{1}{x \ln (10)}}{14 x + 5}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 1} \frac{1}{x \ln (10)} \cdot \frac{1}{14 x + 5}$

Step 5

Multiply $\frac{1}{x \ln (10)}$ by $\frac{1}{14 x + 5}$ .

$lim_{x \to 1} \frac{1}{x \ln (10) (14 x + 5)}$

Step 6

Move the term $\frac{1}{\ln (10)}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{\ln (10)} lim_{x \to 1} \frac{1}{x (14 x + 5)}$

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $1$ .

$\frac{1}{\ln (10)} \cdot \frac{lim_{x \to 1} 1}{lim_{x \to 1} x (14 x + 5)}$

Step 8

Evaluate the limit of $1$ which is constant as $x$ approaches $1$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{lim_{x \to 1} x (14 x + 5)}$

Step 9

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{lim_{x \to 1} x \cdot lim_{x \to 1} 14 x + 5}$

Step 10

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{lim_{x \to 1} x \cdot (lim_{x \to 1} 14 x + lim_{x \to 1} 5)}$

Step 11

Move the term $14$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{lim_{x \to 1} x \cdot (14 lim_{x \to 1} x + lim_{x \to 1} 5)}$

Step 12

Evaluate the limit of $5$ which is constant as $x$ approaches $1$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{lim_{x \to 1} x \cdot (14 lim_{x \to 1} x + 5)}$

Step 13

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 13.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{1 \cdot (14 lim_{x \to 1} x + 5)}$

Step 13.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{1 \cdot (14 \cdot 1 + 5)}$

Step 14

Simplify the answer.

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Step 14.1

Cancel the common factor of $1$ .

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Step 14.1.1

Cancel the common factor.

$\frac{1}{\ln (10)} \cdot \frac{1}{1 \cdot (14 \cdot 1 + 5)}$

Step 14.1.2

Rewrite the expression.

$\frac{1}{\ln (10)} \cdot \frac{1}{14 \cdot 1 + 5}$

Step 14.2

Simplify the denominator.

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Step 14.2.1

Multiply $14$ by $1$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{14 + 5}$

Step 14.2.2

Add $14$ and $5$ .

$\frac{1}{\ln (10)} \cdot \frac{1}{19}$

Step 14.3

Multiply $\frac{1}{\ln (10)}$ by $\frac{1}{19}$ .

$\frac{1}{\ln (10) \cdot 19}$

Step 14.4

Move $19$ to the left of $\ln (10)$ .

$\frac{1}{19 \ln (10)}$