Calculus Examples

$lim_{x \to - 4} \frac{8 x^{2} + 40 x + 32}{x + 4}$

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.

$\frac{lim_{x \to - 4} 8 x^{2} + 40 x + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 4$ .

$\frac{lim_{x \to - 4} 8 x^{2} + lim_{x \to - 4} 40 x + lim_{x \to - 4} 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.2

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

$\frac{8 lim_{x \to - 4} x^{2} + lim_{x \to - 4} 40 x + lim_{x \to - 4} 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.3

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

$\frac{8 {(lim_{x \to - 4} x)}^{2} + lim_{x \to - 4} 40 x + lim_{x \to - 4} 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.4

Move the term $40$ outside of the limit because it is constant with respect to $x$ .

$\frac{8 {(lim_{x \to - 4} x)}^{2} + 40 lim_{x \to - 4} x + lim_{x \to - 4} 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.5

Evaluate the limit of $32$ which is constant as $x$ approaches $- 4$ .

$\frac{8 {(lim_{x \to - 4} x)}^{2} + 40 lim_{x \to - 4} x + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.6

Evaluate the limits by plugging in $- 4$ for all occurrences of $x$ .

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Step 1.1.2.6.1

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{8 {(- 4)}^{2} + 40 lim_{x \to - 4} x + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.6.2

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{8 {(- 4)}^{2} + 40 \cdot - 4 + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.7

Simplify the answer.

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Step 1.1.2.7.1

Simplify each term.

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Step 1.1.2.7.1.1

Raise $- 4$ to the power of $2$ .

$\frac{8 \cdot 16 + 40 \cdot - 4 + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.7.1.2

Multiply $8$ by $16$ .

$\frac{128 + 40 \cdot - 4 + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.7.1.3

Multiply $40$ by $- 4$ .

$\frac{128 - 160 + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.7.2

Subtract $160$ from $128$ .

$\frac{- 32 + 32}{lim_{x \to - 4} x + 4}$

Step 1.1.2.7.3

Add $- 32$ and $32$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 4$ .

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

Step 1.1.3.1.2

Evaluate the limit of $4$ which is constant as $x$ approaches $- 4$ .

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

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{0}{- 4 + 4}$

Step 1.1.3.3

Add $- 4$ and $4$ .

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to - 4} \frac{8 x^{2} + 40 x + 32}{x + 4} = lim_{x \to - 4} \frac{\frac{d}{d x} [8 x^{2} + 40 x + 32]}{\frac{d}{d x} [x + 4]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 4} \frac{\frac{d}{d x} [8 x^{2} + 40 x + 32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.2

By the Sum Rule, the derivative of $8 x^{2} + 40 x + 32$ with respect to $x$ is $\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [40 x] + \frac{d}{d x} [32]$ .

$lim_{x \to - 4} \frac{\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [40 x] + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [8 x^{2}]$ .

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Step 1.3.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to - 4} \frac{8 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [40 x] + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.3.2

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

$lim_{x \to - 4} \frac{8 (2 x) + \frac{d}{d x} [40 x] + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.3.3

Multiply $2$ by $8$ .

$lim_{x \to - 4} \frac{16 x + \frac{d}{d x} [40 x] + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [40 x]$ .

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Step 1.3.4.1

Since $40$ is constant with respect to $x$ , the derivative of $40 x$ with respect to $x$ is $40 \frac{d}{d x} [x]$ .

$lim_{x \to - 4} \frac{16 x + 40 \frac{d}{d x} [x] + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.4.2

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

$lim_{x \to - 4} \frac{16 x + 40 \cdot 1 + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.4.3

Multiply $40$ by $1$ .

$lim_{x \to - 4} \frac{16 x + 40 + \frac{d}{d x} [32]}{\frac{d}{d x} [x + 4]}$

Step 1.3.5

Since $32$ is constant with respect to $x$ , the derivative of $32$ with respect to $x$ is $0$ .

$lim_{x \to - 4} \frac{16 x + 40 + 0}{\frac{d}{d x} [x + 4]}$

Step 1.3.6

Add $16 x + 40$ and $0$ .

$lim_{x \to - 4} \frac{16 x + 40}{\frac{d}{d x} [x + 4]}$

Step 1.3.7

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

$lim_{x \to - 4} \frac{16 x + 40}{\frac{d}{d x} [x] + \frac{d}{d x} [4]}$

Step 1.3.8

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

$lim_{x \to - 4} \frac{16 x + 40}{1 + \frac{d}{d x} [4]}$

Step 1.3.9

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

$lim_{x \to - 4} \frac{16 x + 40}{1 + 0}$

Step 1.3.10

Add $1$ and $0$ .

$lim_{x \to - 4} \frac{16 x + 40}{1}$

Step 1.4

Divide $16 x + 40$ by $1$ .

$lim_{x \to - 4} 16 x + 40$

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 $x$ approaches $- 4$ .

$lim_{x \to - 4} 16 x + lim_{x \to - 4} 40$

Step 2.2

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

$16 lim_{x \to - 4} x + lim_{x \to - 4} 40$

Step 2.3

Evaluate the limit of $40$ which is constant as $x$ approaches $- 4$ .

$16 lim_{x \to - 4} x + 40$

Step 3

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$16 \cdot - 4 + 40$

Step 4

Simplify the answer.

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Step 4.1

Multiply $16$ by $- 4$ .

$- 64 + 40$

Step 4.2

Add $- 64$ and $40$ .

$- 24$