Calculus Examples

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - 8 {(\frac{1}{2})}^{8}}{h}$

Step 1

Evaluate the limit.

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Step 1.1

Simplify the limit argument.

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Step 1.1.1

Combine factors.

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Step 1.1.1.1

Factor out negative.

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (8 {(\frac{1}{2})}^{8})}{h}$

Step 1.1.1.2

Rewrite $8$ as $2^{3}$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} {(\frac{1}{2})}^{8})}{h}$

Step 1.1.1.3

Rewrite $\frac{1}{2}$ as $2^{- 1}$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} {(2^{- 1})}^{8})}{h}$

Step 1.1.1.4

Raise $2$ to the power of $1$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} {({(2^{1})}^{- 1})}^{8})}{h}$

Step 1.1.1.5

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} {(2^{1 \cdot - 1})}^{8})}{h}$

Step 1.1.1.6

Multiply $- 1$ by $1$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} {(2^{- 1})}^{8})}{h}$

Step 1.1.1.7

Multiply the exponents in ${(2^{- 1})}^{8}$ .

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Step 1.1.1.7.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} \cdot 2^{- 1 \cdot 8})}{h}$

Step 1.1.1.7.2

Multiply $- 1$ by $8$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - (2^{3} \cdot 2^{- 8})}{h}$

Step 1.1.1.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - 2^{3 - 8}}{h}$

Step 1.1.1.9

Subtract $8$ from $3$ .

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h)}^{8} - 2^{- 5}}{h}$

Step 1.1.2

Combine terms.

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Step 1.1.2.1

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

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + h \cdot \frac{2}{2})}^{8} - 2^{- 5}}{h}$

Step 1.1.2.2

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

$lim_{h \to 0} \frac{8 {(\frac{1}{2} + \frac{h \cdot 2}{2})}^{8} - 2^{- 5}}{h}$

Step 1.1.2.3

Combine the numerators over the common denominator.

$lim_{h \to 0} \frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} - 2^{- 5}}{h}$

Step 1.2

Simplify terms.

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Step 1.2.1

Simplify the limit argument.

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Step 1.2.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{h \to 0} \frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} - \frac{1}{2^{5}}}{h}$

Step 1.2.1.2

Combine terms.

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Step 1.2.1.2.1

To write $8 {(\frac{1 + h \cdot 2}{2})}^{8}$ as a fraction with a common denominator, multiply by $\frac{2^{5}}{2^{5}}$ .

$lim_{h \to 0} \frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} \cdot \frac{2^{5}}{2^{5}} - \frac{1}{2^{5}}}{h}$

Step 1.2.1.2.2

Combine $8 {(\frac{1 + h \cdot 2}{2})}^{8}$ and $\frac{2^{5}}{2^{5}}$ .

$lim_{h \to 0} \frac{\frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} \cdot 2^{5}}{2^{5}} - \frac{1}{2^{5}}}{h}$

Step 1.2.1.2.3

Combine the numerators over the common denominator.

$lim_{h \to 0} \frac{\frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} \cdot 2^{5} - 1}{2^{5}}}{h}$

Step 1.2.2

Simplify the limit argument.

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Step 1.2.2.1

Multiply the numerator by the reciprocal of the denominator.

$lim_{h \to 0} \frac{8 {(\frac{1 + h \cdot 2}{2})}^{8} \cdot 2^{5} - 1}{2^{5}} \cdot \frac{1}{h}$

Step 1.2.2.2

Combine factors.

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Step 1.2.2.2.1

Rewrite $8$ as $2^{3}$ .

$lim_{h \to 0} \frac{2^{5} \cdot 2^{3} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{2^{5}} \cdot \frac{1}{h}$

Step 1.2.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{h \to 0} \frac{2^{5 + 3} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{2^{5}} \cdot \frac{1}{h}$

Step 1.2.2.2.3

Add $5$ and $3$ .

$lim_{h \to 0} \frac{2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{2^{5}} \cdot \frac{1}{h}$

Step 1.2.2.2.4

Multiply $\frac{2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{2^{5}}$ by $\frac{1}{h}$ .

$lim_{h \to 0} \frac{2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{2^{5} h}$

Step 1.3

Move the term $\frac{1}{2^{5}}$ outside of the limit because it is constant with respect to $h$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{h}$

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.

$\frac{1}{2^{5}} \cdot \frac{lim_{h \to 0} 2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{lim_{h \to 0} h}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot \frac{lim_{h \to 0} 2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.2

Move the term $2^{8}$ outside of the limit because it is constant with respect to $h$ .

$\frac{1}{2^{5}} \cdot \frac{2^{8} lim_{h \to 0} {(\frac{1 + h \cdot 2}{2})}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.3

Move the exponent $8$ from ${(\frac{1 + h \cdot 2}{2})}^{8}$ outside the limit using the Limits Power Rule.

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(lim_{h \to 0} \frac{1 + h \cdot 2}{2})}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.4

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

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} lim_{h \to 0} 1 + h \cdot 2)}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.5

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} (lim_{h \to 0} 1 + lim_{h \to 0} h \cdot 2))}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.6

Evaluate the limit of $1$ which is constant as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} (1 + lim_{h \to 0} h \cdot 2))}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.7

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

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} (1 + 2 lim_{h \to 0} h))}^{8} - lim_{h \to 0} 1}{lim_{h \to 0} h}$

Step 2.1.2.1.8

Evaluate the limit of $1$ which is constant as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} (1 + 2 lim_{h \to 0} h))}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.2

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{1}{2^{5}} \cdot \frac{2^{8} {(\frac{1}{2} (1 + 2 \cdot 0))}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Raise $2$ to the power of $8$ .

$\frac{1}{2^{5}} \cdot \frac{256 {(\frac{1}{2} (1 + 2 \cdot 0))}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.2

Multiply $2$ by $0$ .

$\frac{1}{2^{5}} \cdot \frac{256 {(\frac{1}{2} (1 + 0))}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.3

Add $1$ and $0$ .

$\frac{1}{2^{5}} \cdot \frac{256 {(\frac{1}{2} \cdot 1)}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.4

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

$\frac{1}{2^{5}} \cdot \frac{256 {(\frac{1}{2})}^{8} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.5

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

$\frac{1}{2^{5}} \cdot \frac{256 \frac{1^{8}}{2^{8}} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.6

One to any power is one.

$\frac{1}{2^{5}} \cdot \frac{256 \frac{1}{2^{8}} - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.7

Raise $2$ to the power of $8$ .

$\frac{1}{2^{5}} \cdot \frac{256 (\frac{1}{256}) - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.8

Cancel the common factor of $256$ .

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Step 2.1.2.3.1.8.1

Cancel the common factor.

$\frac{1}{2^{5}} \cdot \frac{256 (\frac{1}{256}) - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.8.2

Rewrite the expression.

$\frac{1}{2^{5}} \cdot \frac{1 - 1 \cdot 1}{lim_{h \to 0} h}$

Step 2.1.2.3.1.9

Multiply $- 1$ by $1$ .

$\frac{1}{2^{5}} \cdot \frac{1 - 1}{lim_{h \to 0} h}$

Step 2.1.2.3.2

Subtract $1$ from $1$ .

$\frac{1}{2^{5}} \cdot \frac{0}{lim_{h \to 0} h}$

Step 2.1.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{1}{2^{5}} \cdot \frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{1}{2^{5}} \cdot \frac{0}{0}$

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.

$lim_{h \to 0} \frac{2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1]}{\frac{d}{d h} [h]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{d}{d h} [2^{8} {(\frac{1 + h \cdot 2}{2})}^{8} - 1]}{\frac{d}{d h} [h]}$

Step 2.3.2

Raise $2$ to the power of $8$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{d}{d h} [256 {(\frac{1 + h \cdot 2}{2})}^{8} - 1]}{\frac{d}{d h} [h]}$

Step 2.3.3

By the Sum Rule, the derivative of $256 {(\frac{1 + h \cdot 2}{2})}^{8} - 1$ with respect to $h$ is $\frac{d}{d h} [256 {(\frac{1 + h \cdot 2}{2})}^{8}] + \frac{d}{d h} [- 1]$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{d}{d h} [256 {(\frac{1 + h \cdot 2}{2})}^{8}] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4

Evaluate $\frac{d}{d h} [256 {(\frac{1 + h \cdot 2}{2})}^{8}]$ .

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Step 2.3.4.1

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{d}{d h} [256 {(\frac{1 + 2 \cdot h}{2})}^{8}] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.2

Since $256$ is constant with respect to $h$ , the derivative of $256 {(\frac{1 + 2 h}{2})}^{8}$ with respect to $h$ is $256 \frac{d}{d h} [{(\frac{1 + 2 h}{2})}^{8}]$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \frac{d}{d h} [{(\frac{1 + 2 h}{2})}^{8}] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.3

Differentiate using the chain rule, which states that $\frac{d}{d h} [f (g (h))]$ is $f' (g (h)) g' (h)$ where $f (h) = h^{8}$ and $g (h) = \frac{1 + 2 h}{2}$ .

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Step 2.3.4.3.1

To apply the Chain Rule, set $u$ as $\frac{1 + 2 h}{2}$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \frac{d}{d u} [u^{8}] \frac{d}{d h} [\frac{1 + 2 h}{2}] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 8$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{d}{d h} [\frac{1 + 2 h}{2}] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.4

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} \frac{d}{d h} [1 + 2 h] + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.5

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} (\frac{d}{d h} [1] + \frac{d}{d h} [2 h]) + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.6

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} (0 + \frac{d}{d h} [2 h]) + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.7

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} (0 + 2 \frac{d}{d h} [h]) + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.8

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} (0 + 2 \cdot 1) + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.9

Multiply $2$ by $1$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} (0 + 2) + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.10

Add $0$ and $2$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{1}{2} \cdot 2 + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.11

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{2}{2} + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.12

Cancel the common factor of $2$ .

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Step 2.3.4.12.1

Cancel the common factor.

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \frac{2}{2} + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.12.2

Rewrite the expression.

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} \cdot 1 + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.13

Multiply $8$ by $1$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{256 \cdot 8 {(\frac{1 + 2 h}{2})}^{7} + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.4.14

Multiply $8$ by $256$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{2048 {(\frac{1 + 2 h}{2})}^{7} + \frac{d}{d h} [- 1]}{\frac{d}{d h} [h]}$

Step 2.3.5

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{2048 {(\frac{1 + 2 h}{2})}^{7} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6

Simplify.

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Step 2.3.6.1

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{2048 \frac{{(1 + 2 h)}^{7}}{2^{7}} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2

Combine terms.

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Step 2.3.6.2.1

Raise $2$ to the power of $7$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{2048 \frac{{(1 + 2 h)}^{7}}{128} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.2

Combine $2048$ and $\frac{{(1 + 2 h)}^{7}}{128}$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{2048 {(1 + 2 h)}^{7}}{128} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.3

Cancel the common factor of $2048$ and $128$ .

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Step 2.3.6.2.3.1

Factor $128$ out of $2048 {(1 + 2 h)}^{7}$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{128 \cdot 16 {(1 + 2 h)}^{7}}{128} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.3.2

Cancel the common factors.

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Step 2.3.6.2.3.2.1

Factor $128$ out of $128$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{128 \cdot 16 {(1 + 2 h)}^{7}}{128 (1)} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.3.2.2

Cancel the common factor.

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{128 \cdot 16 {(1 + 2 h)}^{7}}{128 \cdot 1} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.3.2.3

Rewrite the expression.

$\frac{1}{2^{5}} lim_{h \to 0} \frac{\frac{16 {(1 + 2 h)}^{7}}{1} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.3.2.4

Divide $16 {(1 + 2 h)}^{7}$ by $1$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{16 {(1 + 2 h)}^{7} + 0}{\frac{d}{d h} [h]}$

Step 2.3.6.2.4

Add $16 {(1 + 2 h)}^{7}$ and $0$ .

$\frac{1}{2^{5}} lim_{h \to 0} \frac{16 {(1 + 2 h)}^{7}}{\frac{d}{d h} [h]}$

Step 2.3.7

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

$\frac{1}{2^{5}} lim_{h \to 0} \frac{16 {(1 + 2 h)}^{7}}{1}$

Step 2.4

Divide $16 {(1 + 2 h)}^{7}$ by $1$ .

$\frac{1}{2^{5}} lim_{h \to 0} 16 {(1 + 2 h)}^{7}$

Step 3

Evaluate the limit.

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Step 3.1

Move the term $16$ outside of the limit because it is constant with respect to $h$ .

$\frac{1}{2^{5}} \cdot 16 lim_{h \to 0} {(1 + 2 h)}^{7}$

Step 3.2

Move the exponent $7$ from ${(1 + 2 h)}^{7}$ outside the limit using the Limits Power Rule.

$\frac{1}{2^{5}} \cdot 16 {(lim_{h \to 0} 1 + 2 h)}^{7}$

Step 3.3

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot 16 {(lim_{h \to 0} 1 + lim_{h \to 0} 2 h)}^{7}$

Step 3.4

Evaluate the limit of $1$ which is constant as $h$ approaches $0$ .

$\frac{1}{2^{5}} \cdot 16 {(1 + lim_{h \to 0} 2 h)}^{7}$

Step 3.5

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

$\frac{1}{2^{5}} \cdot 16 {(1 + 2 lim_{h \to 0} h)}^{7}$

Step 4

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{1}{2^{5}} \cdot 16 {(1 + 2 \cdot 0)}^{7}$

Step 5

Simplify the answer.

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Step 5.1

Raise $2$ to the power of $5$ .

$\frac{1}{32} \cdot 16 {(1 + 2 \cdot 0)}^{7}$

Step 5.2

Cancel the common factor of $16$ .

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Step 5.2.1

Factor $16$ out of $32$ .

$\frac{1}{16 (2)} \cdot 16 {(1 + 2 \cdot 0)}^{7}$

Step 5.2.2

Cancel the common factor.

$\frac{1}{16 \cdot 2} \cdot 16 {(1 + 2 \cdot 0)}^{7}$

Step 5.2.3

Rewrite the expression.

$\frac{1}{2} {(1 + 2 \cdot 0)}^{7}$

Step 5.3

Multiply $2$ by $0$ .

$\frac{1}{2} {(1 + 0)}^{7}$

Step 5.4

Add $1$ and $0$ .

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

Step 5.5

One to any power is one.

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

Step 5.6

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

$\frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{2}$

Decimal Form:

$0.5$