Calculus Examples

$lim_{x \to - 2} \frac{{(x + 3)}^{5} - 1}{x^{2} - 4}$

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.

$\frac{lim_{x \to - 2} {(x + 3)}^{5} - 1}{lim_{x \to - 2} x^{2} - 4}$

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 $x$ approaches $- 2$ .

$\frac{lim_{x \to - 2} {(x + 3)}^{5} - lim_{x \to - 2} 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.1.2

Move the exponent $5$ from ${(x + 3)}^{5}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to - 2} x + 3)}^{5} - lim_{x \to - 2} 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{{(lim_{x \to - 2} x + lim_{x \to - 2} 3)}^{5} - lim_{x \to - 2} 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.1.4

Evaluate the limit of $3$ which is constant as $x$ approaches $- 2$ .

$\frac{{(lim_{x \to - 2} x + 3)}^{5} - lim_{x \to - 2} 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.1.5

Evaluate the limit of $1$ which is constant as $x$ approaches $- 2$ .

$\frac{{(lim_{x \to - 2} x + 3)}^{5} - 1 \cdot 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{{(- 2 + 3)}^{5} - 1 \cdot 1}{lim_{x \to - 2} x^{2} - 4}$

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

Add $- 2$ and $3$ .

$\frac{1^{5} - 1 \cdot 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.3.1.2

One to any power is one.

$\frac{1 - 1 \cdot 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.3.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to - 2} x^{2} - 4}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to - 2} x^{2} - 4}$

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 $- 2$ .

$\frac{0}{lim_{x \to - 2} x^{2} - lim_{x \to - 2} 4}$

Step 1.1.3.1.2

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

$\frac{0}{{(lim_{x \to - 2} x)}^{2} - lim_{x \to - 2} 4}$

Step 1.1.3.1.3

Evaluate the limit of $4$ which is constant as $x$ approaches $- 2$ .

$\frac{0}{{(lim_{x \to - 2} x)}^{2} - 1 \cdot 4}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{0}{{(- 2)}^{2} - 1 \cdot 4}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

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

$\frac{0}{4 - 1 \cdot 4}$

Step 1.1.3.3.1.2

Multiply $- 1$ by $4$ .

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

Step 1.1.3.3.2

Subtract $4$ from $4$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to - 2} \frac{{(x + 3)}^{5} - 1}{x^{2} - 4} = lim_{x \to - 2} \frac{\frac{d}{d x} [{(x + 3)}^{5} - 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 2} \frac{\frac{d}{d x} [{(x + 3)}^{5} - 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.2

By the Sum Rule, the derivative of ${(x + 3)}^{5} - 1$ with respect to $x$ is $\frac{d}{d x} [{(x + 3)}^{5}] + \frac{d}{d x} [- 1]$ .

$lim_{x \to - 2} \frac{\frac{d}{d x} [{(x + 3)}^{5}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [{(x + 3)}^{5}]$ .

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Step 1.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x + 3$ .

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Step 1.3.3.1.1

To apply the Chain Rule, set $u$ as $x + 3$ .

$lim_{x \to - 2} \frac{\frac{d}{d u} [u^{5}] \frac{d}{d x} [x + 3] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 5$ .

$lim_{x \to - 2} \frac{5 u^{4} \frac{d}{d x} [x + 3] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.1.3

Replace all occurrences of $u$ with $x + 3$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} \frac{d}{d x} [x + 3] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} (\frac{d}{d x} [x] + \frac{d}{d x} [3]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.3

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

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} (1 + \frac{d}{d x} [3]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} (1 + 0) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.5

Add $1$ and $0$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} \cdot 1 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.3.6

Multiply $5$ by $1$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} + \frac{d}{d x} [- 1]}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.4

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

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4} + 0}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.5

Add $5 {(x + 3)}^{4}$ and $0$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4}}{\frac{d}{d x} [x^{2} - 4]}$

Step 1.3.6

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4}}{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]}$

Step 1.3.7

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

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4}}{2 x + \frac{d}{d x} [- 4]}$

Step 1.3.8

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

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4}}{2 x + 0}$

Step 1.3.9

Add $2 x$ and $0$ .

$lim_{x \to - 2} \frac{5 {(x + 3)}^{4}}{2 x}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{5}{2} lim_{x \to - 2} \frac{{(x + 3)}^{4}}{x}$

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 2$ .

$\frac{5}{2} \cdot \frac{lim_{x \to - 2} {(x + 3)}^{4}}{lim_{x \to - 2} x}$

Step 2.3

Move the exponent $4$ from ${(x + 3)}^{4}$ outside the limit using the Limits Power Rule.

$\frac{5}{2} \cdot \frac{{(lim_{x \to - 2} x + 3)}^{4}}{lim_{x \to - 2} x}$

Step 2.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{5}{2} \cdot \frac{{(lim_{x \to - 2} x + lim_{x \to - 2} 3)}^{4}}{lim_{x \to - 2} x}$

Step 2.5

Evaluate the limit of $3$ which is constant as $x$ approaches $- 2$ .

$\frac{5}{2} \cdot \frac{{(lim_{x \to - 2} x + 3)}^{4}}{lim_{x \to - 2} x}$

Step 3

Evaluate the limits by plugging in $- 2$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{5}{2} \cdot \frac{{(- 2 + 3)}^{4}}{lim_{x \to - 2} x}$

Step 3.2

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{5}{2} \cdot \frac{{(- 2 + 3)}^{4}}{- 2}$

Step 4

Simplify the answer.

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Step 4.1

Combine.

$\frac{5 {(- 2 + 3)}^{4}}{2 \cdot - 2}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Add $- 2$ and $3$ .

$\frac{5 \cdot 1^{4}}{2 \cdot - 2}$

Step 4.2.2

One to any power is one.

$\frac{5 \cdot 1}{2 \cdot - 2}$

Step 4.3

Multiply $2$ by $- 2$ .

$\frac{5 \cdot 1}{- 4}$

Step 4.4

Multiply $5$ by $1$ .

$\frac{5}{- 4}$

Step 4.5

Move the negative in front of the fraction.

$- \frac{5}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$- \frac{5}{4}$

Decimal Form:

$- 1.25$