Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

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

Step 1.2.2

Move the limit inside the logarithm.

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

Step 1.2.3

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.2.4

Simplify terms.

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Step 1.2.4.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.2.4.2

Simplify the answer.

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Step 1.2.4.2.1

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

$\frac{2 \cdot 0}{lim_{x \to - 1} e^{3 x + 3} - 1}$

Step 1.2.4.2.2

Multiply $2$ by $0$ .

$\frac{0}{lim_{x \to - 1} e^{3 x + 3} - 1}$

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 $- 1$ .

$\frac{0}{lim_{x \to - 1} e^{3 x + 3} - lim_{x \to - 1} 1}$

Step 1.3.1.2

Move the limit into the exponent.

$\frac{0}{e^{lim_{x \to - 1} 3 x + 3} - lim_{x \to - 1} 1}$

Step 1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{0}{e^{lim_{x \to - 1} 3 x + lim_{x \to - 1} 3} - lim_{x \to - 1} 1}$

Step 1.3.1.4

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{e^{3 lim_{x \to - 1} x + lim_{x \to - 1} 3} - lim_{x \to - 1} 1}$

Step 1.3.1.5

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

$\frac{0}{e^{3 lim_{x \to - 1} x + 3} - lim_{x \to - 1} 1}$

Step 1.3.1.6

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

$\frac{0}{e^{3 lim_{x \to - 1} x + 3} - 1 \cdot 1}$

Step 1.3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{0}{e^{3 \cdot - 1 + 3} - 1 \cdot 1}$

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

Multiply $3$ by $- 1$ .

$\frac{0}{e^{- 3 + 3} - 1 \cdot 1}$

Step 1.3.3.1.2

Add $- 3$ and $3$ .

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

Step 1.3.3.1.3

Anything raised to $0$ is $1$ .

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

Step 1.3.3.1.4

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 \ln (- x)$ with respect to $x$ is $2 \frac{d}{d x} [\ln (- x)]$ .

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

Step 3.3

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 $g (x) = - x$ .

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Step 3.3.1

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

$lim_{x \to - 1} \frac{2 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [- x])}{\frac{d}{d x} [e^{3 x + 3} - 1]}$

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

Cancel the common factor of $1$ and $- 1$ .

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Step 3.4.1

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

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

Step 3.4.2

Move the negative in front of the fraction.

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

Step 3.5

Multiply $- 1$ by $2$ .

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

Step 3.6

Combine $\frac{1}{x}$ and $- 2$ .

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

Step 3.7

Move the negative in front of the fraction.

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

Step 3.8

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

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

Step 3.9

Multiply $- 1$ by $- 1$ .

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

Step 3.10

Multiply $\frac{2}{x}$ by $1$ .

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

Step 3.11

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

$lim_{x \to - 1} \frac{\frac{2}{x} \cdot 1}{\frac{d}{d x} [e^{3 x + 3} - 1]}$

Step 3.12

Multiply $\frac{2}{x}$ by $1$ .

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

Step 3.13

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

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

Step 3.14

Evaluate $\frac{d}{d x} [e^{3 x + 3}]$ .

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Step 3.14.1

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 $g (x) = 3 x + 3$ .

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Step 3.14.1.1

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

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

Step 3.14.1.2

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

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

Step 3.14.1.3

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

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

Step 3.14.2

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

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

Step 3.14.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 3.14.4

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

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

Step 3.14.5

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

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

Step 3.14.6

Multiply $3$ by $1$ .

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

Step 3.14.7

Add $3$ and $0$ .

$lim_{x \to - 1} \frac{\frac{2}{x}}{e^{3 x + 3} \cdot 3 + \frac{d}{d x} [- 1]}$

Step 3.14.8

Move $3$ to the left of $e^{3 x + 3}$ .

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

Step 3.15

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

$lim_{x \to - 1} \frac{\frac{2}{x}}{3 e^{3 x + 3} + 0}$

Step 3.16

Add $3 e^{3 x + 3}$ and $0$ .

$lim_{x \to - 1} \frac{\frac{2}{x}}{3 e^{3 x + 3}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 1} \frac{2}{x} \cdot \frac{1}{3 e^{3 x + 3}}$

Step 5

Multiply $\frac{2}{x}$ by $\frac{1}{3 e^{3 x + 3}}$ .

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

Step 6

Move the term $\frac{2}{3}$ outside of the limit because it is constant with respect to $x$ .

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

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 1$ .

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

Step 8

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

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

Step 9

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{2}{3} \cdot \frac{1}{lim_{x \to - 1} x \cdot lim_{x \to - 1} e^{3 x + 3}}$

Step 10

Move the limit into the exponent.

$\frac{2}{3} \cdot \frac{1}{lim_{x \to - 1} x \cdot e^{lim_{x \to - 1} 3 x + 3}}$

Step 11

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{2}{3} \cdot \frac{1}{lim_{x \to - 1} x \cdot e^{lim_{x \to - 1} 3 x + lim_{x \to - 1} 3}}$

Step 12

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{2}{3} \cdot \frac{1}{lim_{x \to - 1} x \cdot e^{3 lim_{x \to - 1} x + lim_{x \to - 1} 3}}$

Step 13

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

$\frac{2}{3} \cdot \frac{1}{lim_{x \to - 1} x \cdot e^{3 lim_{x \to - 1} x + 3}}$

Step 14

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 14.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{2}{3} \cdot \frac{1}{- 1 \cdot e^{3 lim_{x \to - 1} x + 3}}$

Step 14.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{2}{3} \cdot \frac{1}{- 1 \cdot e^{3 \cdot - 1 + 3}}$

Step 15

Simplify the answer.

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Step 15.1

Combine.

$\frac{2 \cdot 1}{3 (- 1 \cdot e^{3 \cdot - 1 + 3})}$

Step 15.2

Cancel the common factor of $1$ and $- 1$ .

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Step 15.2.1

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

$\frac{2 \cdot (- 1 (- 1))}{3 (- 1 \cdot e^{3 \cdot - 1 + 3})}$

Step 15.2.2

Move the negative in front of the fraction.

$- \frac{2}{3 (e^{3 \cdot - 1 + 3})}$

Step 15.3

Simplify the denominator.

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Step 15.3.1

Multiply $3$ by $- 1$ .

$- \frac{2}{3 e^{- 3 + 3}}$

Step 15.3.2

Add $- 3$ and $3$ .

$- \frac{2}{3 e^{0}}$

Step 15.3.3

Anything raised to $0$ is $1$ .

$- \frac{2}{3 \cdot 1}$

Step 15.4

Multiply $3$ by $1$ .

$- \frac{2}{3}$