Calculus Examples

$lim_{x \to - 2} \frac{\ln (- 1 - x)}{e^{x + 2} - 1}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to - 2} \ln (- 1 - x)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Move the limit inside the logarithm.

$\frac{\ln (lim_{x \to - 2} - 1 - x)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.2

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

$\frac{\ln (- lim_{x \to - 2} 1 - lim_{x \to - 2} x)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.3

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

$\frac{\ln (- 1 \cdot 1 - lim_{x \to - 2} x)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.4

Simplify terms.

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Step 1.2.4.1

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

$\frac{\ln (- 1 \cdot 1 + 2)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.4.2

Simplify the answer.

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Step 1.2.4.2.1

Multiply $- 1$ by $1$ .

$\frac{\ln (- 1 + 2)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.4.2.2

Add $- 1$ and $2$ .

$\frac{\ln (1)}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.2.4.2.3

The natural logarithm of $1$ is $0$ .

$\frac{0}{lim_{x \to - 2} e^{x + 2} - 1}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

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

$\frac{0}{lim_{x \to - 2} e^{x + 2} - lim_{x \to - 2} 1}$

Step 1.3.1.2

Move the limit into the exponent.

$\frac{0}{e^{lim_{x \to - 2} x + 2} - lim_{x \to - 2} 1}$

Step 1.3.1.3

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

$\frac{0}{e^{lim_{x \to - 2} x + lim_{x \to - 2} 2} - lim_{x \to - 2} 1}$

Step 1.3.1.4

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

$\frac{0}{e^{lim_{x \to - 2} x + 2} - lim_{x \to - 2} 1}$

Step 1.3.1.5

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

$\frac{0}{e^{lim_{x \to - 2} x + 2} - 1 \cdot 1}$

Step 1.3.2

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

$\frac{0}{e^{- 2 + 2} - 1 \cdot 1}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Add $- 2$ and $2$ .

$\frac{0}{e^{0} - 1 \cdot 1}$

Step 1.3.3.1.2

Anything raised to $0$ is $1$ .

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

Step 1.3.3.1.3

Multiply $- 1$ by $1$ .

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

Step 1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{x \to - 2} \frac{\ln (- 1 - x)}{e^{x + 2} - 1} = lim_{x \to - 2} \frac{\frac{d}{d x} [\ln (- 1 - x)]}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - 2} \frac{\frac{d}{d x} [\ln (- 1 - x)]}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = - 1 - x$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $- 1 - x$ .

$lim_{x \to - 2} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [- 1 - x]}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

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

Step 3.2.3

Replace all occurrences of $u$ with $- 1 - x$ .

$lim_{x \to - 2} \frac{\frac{1}{- 1 - x} \frac{d}{d x} [- 1 - x]}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Add $0$ and $\frac{d}{d x} [- x]$ .

$lim_{x \to - 2} \frac{\frac{1}{- 1 - x} \frac{d}{d x} [- x]}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

Multiply $- 1$ by $1$ .

$lim_{x \to - 2} \frac{\frac{1}{- 1 - x} \cdot - 1}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.9

Combine $\frac{1}{- 1 - x}$ and $- 1$ .

$lim_{x \to - 2} \frac{\frac{- 1}{- 1 - x}}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.10

Move the negative in front of the fraction.

$lim_{x \to - 2} \frac{- \frac{1}{- 1 - x}}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.11

Simplify.

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Step 3.11.1

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 3.11.2

Factor $- 1$ out of $- x$ .

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

Step 3.11.3

Factor $- 1$ out of $- 1 (1) - (x)$ .

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

Step 3.11.4

Move the negative in front of the fraction.

$lim_{x \to - 2} \frac{- - \frac{1}{1 + x}}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.11.5

Multiply $- 1$ by $- 1$ .

$lim_{x \to - 2} \frac{1 \frac{1}{1 + x}}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.11.6

Multiply $\frac{1}{1 + x}$ by $1$ .

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{\frac{d}{d x} [e^{x + 2} - 1]}$

Step 3.12

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

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{\frac{d}{d x} [e^{x + 2}] + \frac{d}{d x} [- 1]}$

Step 3.13

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

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Step 3.13.1

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

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Step 3.13.1.1

To apply the Chain Rule, set $u$ as $x + 2$ .

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

Step 3.13.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 3.13.1.3

Replace all occurrences of $u$ with $x + 2$ .

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{e^{x + 2} \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 1]}$

Step 3.13.2

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

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

Step 3.13.3

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

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

Step 3.13.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 3.13.5

Add $1$ and $0$ .

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{e^{x + 2} \cdot 1 + \frac{d}{d x} [- 1]}$

Step 3.13.6

Multiply $e^{x + 2}$ by $1$ .

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{e^{x + 2} + \frac{d}{d x} [- 1]}$

Step 3.14

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

$lim_{x \to - 2} \frac{\frac{1}{1 + x}}{e^{x + 2} + 0}$

Step 3.15

Add $e^{x + 2}$ and $0$ .

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

Multiply $\frac{1}{1 + x}$ by $\frac{1}{e^{x + 2}}$ .

$lim_{x \to - 2} \frac{1}{(1 + x) e^{x + 2}}$

Step 6

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

$\frac{lim_{x \to - 2} 1}{lim_{x \to - 2} (1 + x) e^{x + 2}}$

Step 7

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

$\frac{1}{lim_{x \to - 2} (1 + x) e^{x + 2}}$

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Move the limit into the exponent.

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

Step 12

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

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

Step 13

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

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

$\frac{1}{(1 - 2) \cdot e^{- 2 + 2}}$

Step 15

Simplify the answer.

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Step 15.1

Simplify the denominator.

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Step 15.1.1

Subtract $2$ from $1$ .

$\frac{1}{- 1 e^{- 2 + 2}}$

Step 15.1.2

Add $- 2$ and $2$ .

$\frac{1}{- 1 e^{0}}$

Step 15.1.3

Anything raised to $0$ is $1$ .

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

Step 15.2

Multiply $- 1$ by $1$ .

$\frac{1}{- 1}$

Step 15.3

Divide $1$ by $- 1$ .

$- 1$