Calculus Examples

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\ln (e^{x} + x)}{x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

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.

$lim_{x \to 0} \ln (y) = \frac{lim_{x \to 0} \ln (e^{x} + x)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Move the limit inside the logarithm.

$lim_{x \to 0} \ln (y) = \frac{\ln (lim_{x \to 0} e^{x} + x)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.2

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

$lim_{x \to 0} \ln (y) = \frac{\ln (lim_{x \to 0} e^{x} + lim_{x \to 0} x)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.3

Move the limit into the exponent.

$lim_{x \to 0} \ln (y) = \frac{\ln (e^{lim_{x \to 0} x} + lim_{x \to 0} x)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.4

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.2.4.1

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

$lim_{x \to 0} \ln (y) = \frac{\ln (e^{0} + lim_{x \to 0} x)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.4.2

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

$lim_{x \to 0} \ln (y) = \frac{\ln (e^{0} + 0)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.5

Simplify the answer.

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Step 1.2.5.1

Anything raised to $0$ is $1$ .

$lim_{x \to 0} \ln (y) = \frac{\ln (1 + 0)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.5.2

Add $1$ and $0$ .

$lim_{x \to 0} \ln (y) = \frac{\ln (1)}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.2.5.3

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

$lim_{x \to 0} \ln (y) = \frac{0}{lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 1.3

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

$lim_{x \to 0} \ln (y) = \frac{0}{0} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

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.

$lim_{x \to 0} \frac{\ln (e^{x} + x)}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [\ln (e^{x} + x)]}{\frac{d}{d x} [x]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{d}{d x} [\ln (e^{x} + x)]}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

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 $g (x) = e^{x} + x$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $e^{x} + x$ .

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [e^{x} + x]}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.2.2

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{1}{u} \frac{d}{d x} [e^{x} + x]}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.2.3

Replace all occurrences of $u$ with $e^{x} + x$ .

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{1}{e^{x} + x} \frac{d}{d x} [e^{x} + x]}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.3

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{1}{e^{x} + x} (\frac{d}{d x} [e^{x}] + \frac{d}{d x} [x])}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.4

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{1}{e^{x} + x} (e^{x} + \frac{d}{d x} [x])}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.5

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{1}{e^{x} + x} (e^{x} + 1)}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.6

Simplify.

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Step 3.6.1

Reorder the factors of $\frac{1}{e^{x} + x} (e^{x} + 1)$ .

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{(e^{x} + 1) \frac{1}{e^{x} + x}}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.6.2

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{e^{x} + 1}{e^{x} + x}}{\frac{d}{d x} [x]} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 3.7

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{\frac{e^{x} + 1}{e^{x} + x}}{1} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x} \cdot 1 = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 5

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

$lim_{x \to 0} \ln (y) = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 6

Evaluate the limit of $\ln (y)$ which is constant as $x$ approaches $0$ .

$\ln (y) = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $0$ .

$\ln (y) = \frac{lim_{x \to 0} e^{x} + 1}{lim_{x \to 0} e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 8

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

$\ln (y) = \frac{lim_{x \to 0} e^{x} + lim_{x \to 0} 1}{lim_{x \to 0} e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 9

Move the limit into the exponent.

$\ln (y) = \frac{e^{lim_{x \to 0} x} + lim_{x \to 0} 1}{lim_{x \to 0} e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 10

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{lim_{x \to 0} e^{x} + x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 11

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

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{lim_{x \to 0} e^{x} + lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 12

Move the limit into the exponent.

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = lim_{x \to 0} \frac{e^{x} + 1}{e^{x} + x}$

Step 13

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $0$ .

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{lim_{x \to 0} e^{x} + 1}{lim_{x \to 0} e^{x} + x}$

Step 14

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

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{lim_{x \to 0} e^{x} + lim_{x \to 0} 1}{lim_{x \to 0} e^{x} + x}$

Step 15

Move the limit into the exponent.

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + lim_{x \to 0} 1}{lim_{x \to 0} e^{x} + x}$

Step 16

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + 1}{lim_{x \to 0} e^{x} + x}$

Step 17

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

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + 1}{lim_{x \to 0} e^{x} + lim_{x \to 0} x}$

Step 18

Move the limit into the exponent.

$\ln (y) = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x}$

Step 19

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 19.1

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

$\ln (y) = \frac{e^{0} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x}$

Step 19.2

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

$\ln (y) = \frac{e^{0} + 1}{e^{0} + lim_{x \to 0} x} = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x}$

Step 19.3

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

$\ln (y) = \frac{e^{0} + 1}{e^{0} + 0} = \frac{e^{lim_{x \to 0} x} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x}$

Step 19.4

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

$\ln (y) = \frac{e^{0} + 1}{e^{0} + 0} = \frac{e^{0} + 1}{e^{lim_{x \to 0} x} + lim_{x \to 0} x}$

Step 19.5

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

$\ln (y) = \frac{e^{0} + 1}{e^{0} + 0} = \frac{e^{0} + 1}{e^{0} + lim_{x \to 0} x}$

Step 19.6

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

$\ln (y) = \frac{e^{0} + 1}{e^{0} + 0} = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20

Simplify the answer.

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Step 20.1

Simplify the numerator.

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Step 20.1.1

Anything raised to $0$ is $1$ .

$\ln (y) = \frac{1 + 1}{e^{0} + 0} = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20.1.2

Add $1$ and $1$ .

$\ln (y) = \frac{2}{e^{0} + 0} = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20.2

Simplify the denominator.

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Step 20.2.1

Anything raised to $0$ is $1$ .

$\ln (y) = \frac{2}{1 + 0} = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20.2.2

Add $1$ and $0$ .

$\ln (y) = \frac{2}{1} = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20.3

Divide $2$ by $1$ .

$\ln (y) = 2 = \frac{e^{0} + 1}{e^{0} + 0}$

Step 20.4

Simplify the numerator.

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Step 20.4.1

Anything raised to $0$ is $1$ .

$\ln (y) = 2 = \frac{1 + 1}{e^{0} + 0}$

Step 20.4.2

Add $1$ and $1$ .

$\ln (y) = 2 = \frac{2}{e^{0} + 0}$

Step 20.5

Simplify the denominator.

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Step 20.5.1

Anything raised to $0$ is $1$ .

$\ln (y) = 2 = \frac{2}{1 + 0}$

Step 20.5.2

Add $1$ and $0$ .

$\ln (y) = 2 = \frac{2}{1}$

Step 20.6

Divide $2$ by $1$ .

$\ln (y) = 2 = 2$