Calculus Examples

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

Step 1

Evaluate the limit.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Evaluate the limit of $2$ which is constant as $h$ approaches $0$ .

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

Step 1.5

Evaluate the limit of $2^{- 1}$ which is constant as $h$ approaches $0$ .

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

Step 2

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

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

Step 3

Simplify the answer.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Add $2$ and $0$ .

$\frac{2^{- 1} - 2^{- 1}}{h}$

Step 3.1.2

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

$\frac{\frac{1}{2} - 2^{- 1}}{h}$

Step 3.1.3

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

$\frac{\frac{1}{2} - \frac{1}{2}}{h}$

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{\frac{1 - 1}{2}}{h}$

Step 3.1.5

Subtract $1$ from $1$ .

$\frac{\frac{0}{2}}{h}$

Step 3.1.6

Divide $0$ by $2$ .

$\frac{0}{h}$

Step 3.2

Divide $0$ by $h$ .

$0$