Calculus Examples

$lim_{h \to 0} \frac{{(- 8 + h)}^{2} - 64}{h}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{h \to 0} {(- 8 + h)}^{2} - 64}{lim_{h \to 0} h}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

$\frac{lim_{h \to 0} {(- 8 + h)}^{2} - lim_{h \to 0} 64}{lim_{h \to 0} h}$

Step 1.1.2.1.2

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

$\frac{{(lim_{h \to 0} - 8 + h)}^{2} - lim_{h \to 0} 64}{lim_{h \to 0} h}$

Step 1.1.2.1.3

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

$\frac{{(- lim_{h \to 0} 8 + lim_{h \to 0} h)}^{2} - lim_{h \to 0} 64}{lim_{h \to 0} h}$

Step 1.1.2.1.4

Evaluate the limit of $8$ which is constant as $h$ approaches $0$ .

$\frac{{(- 1 \cdot 8 + lim_{h \to 0} h)}^{2} - lim_{h \to 0} 64}{lim_{h \to 0} h}$

Step 1.1.2.1.5

Evaluate the limit of $64$ which is constant as $h$ approaches $0$ .

$\frac{{(- 1 \cdot 8 + lim_{h \to 0} h)}^{2} - 1 \cdot 64}{lim_{h \to 0} h}$

Step 1.1.2.2

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

$\frac{{(- 1 \cdot 8 + 0)}^{2} - 1 \cdot 64}{lim_{h \to 0} h}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $- 1$ by $8$ .

$\frac{{(- 8 + 0)}^{2} - 1 \cdot 64}{lim_{h \to 0} h}$

Step 1.1.2.3.1.2

Add $- 8$ and $0$ .

$\frac{{(- 8)}^{2} - 1 \cdot 64}{lim_{h \to 0} h}$

Step 1.1.2.3.1.3

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

$\frac{64 - 1 \cdot 64}{lim_{h \to 0} h}$

Step 1.1.2.3.1.4

Multiply $- 1$ by $64$ .

$\frac{64 - 64}{lim_{h \to 0} h}$

Step 1.1.2.3.2

Subtract $64$ from $64$ .

$\frac{0}{lim_{h \to 0} h}$

Step 1.1.3

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

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{h \to 0} \frac{{(- 8 + h)}^{2} - 64}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [{(- 8 + h)}^{2} - 64]}{\frac{d}{d h} [h]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{h \to 0} \frac{\frac{d}{d h} [{(- 8 + h)}^{2} - 64]}{\frac{d}{d h} [h]}$

Step 1.3.2

Rewrite ${(- 8 + h)}^{2}$ as $(- 8 + h) (- 8 + h)$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [(- 8 + h) (- 8 + h) - 64]}{\frac{d}{d h} [h]}$

Step 1.3.3

Expand $(- 8 + h) (- 8 + h)$ using the FOIL Method.

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Step 1.3.3.1

Apply the distributive property.

$lim_{h \to 0} \frac{\frac{d}{d h} [- 8 (- 8 + h) + h (- 8 + h) - 64]}{\frac{d}{d h} [h]}$

Step 1.3.3.2

Apply the distributive property.

$lim_{h \to 0} \frac{\frac{d}{d h} [- 8 \cdot - 8 - 8 h + h (- 8 + h) - 64]}{\frac{d}{d h} [h]}$

Step 1.3.3.3

Apply the distributive property.

$lim_{h \to 0} \frac{\frac{d}{d h} [- 8 \cdot - 8 - 8 h + h \cdot - 8 + h \cdot h - 64]}{\frac{d}{d h} [h]}$

Step 1.3.4

Simplify and combine like terms.

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Step 1.3.4.1

Simplify each term.

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Step 1.3.4.1.1

Multiply $- 8$ by $- 8$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [64 - 8 h + h \cdot - 8 + h \cdot h - 64]}{\frac{d}{d h} [h]}$

Step 1.3.4.1.2

Move $- 8$ to the left of $h$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [64 - 8 h - 8 \cdot h + h \cdot h - 64]}{\frac{d}{d h} [h]}$

Step 1.3.4.1.3

Multiply $h$ by $h$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [64 - 8 h - 8 h + h^{2} - 64]}{\frac{d}{d h} [h]}$

Step 1.3.4.2

Subtract $8 h$ from $- 8 h$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [64 - 16 h + h^{2} - 64]}{\frac{d}{d h} [h]}$

Step 1.3.5

By the Sum Rule, the derivative of $64 - 16 h + h^{2} - 64$ with respect to $h$ is $\frac{d}{d h} [64] + \frac{d}{d h} [- 16 h] + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [64] + \frac{d}{d h} [- 16 h] + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.6

Since $64$ is constant with respect to $h$ , the derivative of $64$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{0 + \frac{d}{d h} [- 16 h] + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.7

Evaluate $\frac{d}{d h} [- 16 h]$ .

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Step 1.3.7.1

Since $- 16$ is constant with respect to $h$ , the derivative of $- 16 h$ with respect to $h$ is $- 16 \frac{d}{d h} [h]$ .

$lim_{h \to 0} \frac{0 - 16 \frac{d}{d h} [h] + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.7.2

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

$lim_{h \to 0} \frac{0 - 16 \cdot 1 + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.7.3

Multiply $- 16$ by $1$ .

$lim_{h \to 0} \frac{0 - 16 + \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.8

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

$lim_{h \to 0} \frac{0 - 16 + 2 h + \frac{d}{d h} [- 64]}{\frac{d}{d h} [h]}$

Step 1.3.9

Since $- 64$ is constant with respect to $h$ , the derivative of $- 64$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{0 - 16 + 2 h + 0}{\frac{d}{d h} [h]}$

Step 1.3.10

Simplify.

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Step 1.3.10.1

Combine terms.

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Step 1.3.10.1.1

Subtract $16$ from $0$ .

$lim_{h \to 0} \frac{- 16 + 2 h + 0}{\frac{d}{d h} [h]}$

Step 1.3.10.1.2

Add $- 16 + 2 h$ and $0$ .

$lim_{h \to 0} \frac{- 16 + 2 h}{\frac{d}{d h} [h]}$

Step 1.3.10.2

Reorder terms.

$lim_{h \to 0} \frac{2 h - 16}{\frac{d}{d h} [h]}$

Step 1.3.11

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

$lim_{h \to 0} \frac{2 h - 16}{1}$

Step 1.4

Divide $2 h - 16$ by $1$ .

$lim_{h \to 0} 2 h - 16$

Step 2

Evaluate the limit.

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Step 2.1

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

$lim_{h \to 0} 2 h - lim_{h \to 0} 16$

Step 2.2

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

$2 lim_{h \to 0} h - lim_{h \to 0} 16$

Step 2.3

Evaluate the limit of $16$ which is constant as $h$ approaches $0$ .

$2 lim_{h \to 0} h - 1 \cdot 16$

Step 3

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

$2 \cdot 0 - 1 \cdot 16$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $2$ by $0$ .

$0 - 1 \cdot 16$

Step 4.1.2

Multiply $- 1$ by $16$ .

$0 - 16$

Step 4.2

Subtract $16$ from $0$ .

$- 16$