Calculus Examples

$lim_{x \to 0} \frac{1}{\ln (x + 1)} - \frac{1}{x}$

Step 1

Combine terms.

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Step 1.1

To write $\frac{1}{\ln (x + 1)}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$lim_{x \to 0} \frac{1}{\ln (x + 1)} \cdot \frac{x}{x} - \frac{1}{x}$

Step 1.2

To write $- \frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{\ln (x + 1)}{\ln (x + 1)}$ .

$lim_{x \to 0} \frac{1}{\ln (x + 1)} \cdot \frac{x}{x} - \frac{1}{x} \cdot \frac{\ln (x + 1)}{\ln (x + 1)}$

Step 1.3

Write each expression with a common denominator of $\ln (x + 1) x$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{1}{\ln (x + 1)}$ by $\frac{x}{x}$ .

$lim_{x \to 0} \frac{x}{\ln (x + 1) x} - \frac{1}{x} \cdot \frac{\ln (x + 1)}{\ln (x + 1)}$

Step 1.3.2

Multiply $\frac{1}{x}$ by $\frac{\ln (x + 1)}{\ln (x + 1)}$ .

$lim_{x \to 0} \frac{x}{\ln (x + 1) x} - \frac{\ln (x + 1)}{x \ln (x + 1)}$

Step 1.3.3

Reorder the factors of $\ln (x + 1) x$ .

$lim_{x \to 0} \frac{x}{x \ln (x + 1)} - \frac{\ln (x + 1)}{x \ln (x + 1)}$

Step 1.4

Combine the numerators over the common denominator.

$lim_{x \to 0} \frac{x - \ln (x + 1)}{x \ln (x + 1)}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

$\frac{lim_{x \to 0} x - \ln (x + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

$\frac{lim_{x \to 0} x - lim_{x \to 0} \ln (x + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.2

Move the limit inside the logarithm.

$\frac{lim_{x \to 0} x - \ln (lim_{x \to 0} x + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.3

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

$\frac{lim_{x \to 0} x - \ln (lim_{x \to 0} x + lim_{x \to 0} 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.4

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

$\frac{lim_{x \to 0} x - \ln (lim_{x \to 0} x + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.5

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

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Step 2.1.2.5.1

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

$\frac{0 - \ln (lim_{x \to 0} x + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.5.2

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

$\frac{0 - \ln (0 + 1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.6

Simplify the answer.

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Step 2.1.2.6.1

Simplify each term.

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Step 2.1.2.6.1.1

Add $0$ and $1$ .

$\frac{0 - \ln (1)}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.6.1.2

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

$\frac{0 - 0}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.6.1.3

Multiply $- 1$ by $0$ .

$\frac{0 + 0}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.2.6.2

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 0} x \ln (x + 1)}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

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

$\frac{0}{lim_{x \to 0} x \cdot lim_{x \to 0} \ln (x + 1)}$

Step 2.1.3.2

Move the limit inside the logarithm.

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1)}$

Step 2.1.3.3

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + lim_{x \to 0} 1)}$

Step 2.1.3.4

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1)}$

Step 2.1.3.5

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

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Step 2.1.3.5.1

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

$\frac{0}{0 \cdot \ln (lim_{x \to 0} x + 1)}$

Step 2.1.3.5.2

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

$\frac{0}{0 \cdot \ln (0 + 1)}$

Step 2.1.3.6

Simplify the answer.

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Step 2.1.3.6.1

Add $0$ and $1$ .

$\frac{0}{0 \cdot \ln (1)}$

Step 2.1.3.6.2

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

$\frac{0}{0 \cdot 0}$

Step 2.1.3.6.3

Multiply $0$ by $0$ .

$\frac{0}{0}$

Step 2.1.3.6.4

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

Undefined

$\frac{0}{0}$

Step 2.1.3.7

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

Undefined

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Evaluate $\frac{d}{d x} [- \ln (x + 1)]$ .

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Step 2.3.4.1

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

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

Step 2.3.4.2

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 $g (x) = x + 1$ .

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Step 2.3.4.2.1

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

$lim_{x \to 0} \frac{1 - (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x + 1])}{\frac{d}{d x} [x \ln (x + 1)]}$

Step 2.3.4.2.2

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

$lim_{x \to 0} \frac{1 - (\frac{1}{u} \frac{d}{d x} [x + 1])}{\frac{d}{d x} [x \ln (x + 1)]}$

Step 2.3.4.2.3

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

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

Step 2.3.4.3

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

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

Step 2.3.4.4

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

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

Step 2.3.4.5

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

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

Step 2.3.4.6

Add $1$ and $0$ .

$lim_{x \to 0} \frac{1 - (\frac{1}{x + 1} \cdot 1)}{\frac{d}{d x} [x \ln (x + 1)]}$

Step 2.3.4.7

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

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

Step 2.3.5

Combine terms.

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Step 2.3.5.1

Write $1$ as a fraction with a common denominator.

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

Step 2.3.5.2

Combine the numerators over the common denominator.

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

Step 2.3.5.3

Subtract $1$ from $1$ .

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

Step 2.3.5.4

Add $x$ and $0$ .

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

Step 2.3.6

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (x + 1)$ .

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

Step 2.3.7

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 $g (x) = x + 1$ .

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Step 2.3.7.1

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

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

Step 2.3.7.2

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

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

Step 2.3.7.3

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

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

Step 2.3.8

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

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

Step 2.3.9

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

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

Step 2.3.10

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

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

Step 2.3.11

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

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

Step 2.3.12

Add $1$ and $0$ .

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x}{x + 1} \cdot 1 + \ln (x + 1) \frac{d}{d x} [x]}$

Step 2.3.13

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

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

Step 2.3.14

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

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x}{x + 1} + \ln (x + 1) \cdot 1}$

Step 2.3.15

Multiply $\ln (x + 1)$ by $1$ .

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x}{x + 1} + \ln (x + 1)}$

Step 2.3.16

To write $\ln (x + 1)$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x}{x + 1} + \frac{\ln (x + 1) (x + 1)}{x + 1}}$

Step 2.3.17

Combine the numerators over the common denominator.

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x + \ln (x + 1) (x + 1)}{x + 1}}$

Step 2.3.18

Simplify.

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Step 2.3.18.1

Simplify the numerator.

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Step 2.3.18.1.1

Simplify each term.

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Step 2.3.18.1.1.1

Apply the distributive property.

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x + \ln (x + 1) x + \ln (x + 1) \cdot 1}{x + 1}}$

Step 2.3.18.1.1.2

Multiply $\ln (x + 1)$ by $1$ .

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x + \ln (x + 1) x + \ln (x + 1)}{x + 1}}$

Step 2.3.18.1.2

Reorder factors in $x + \ln (x + 1) x + \ln (x + 1)$ .

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x + x \ln (x + 1) + \ln (x + 1)}{x + 1}}$

Step 2.3.18.2

Reorder terms.

$lim_{x \to 0} \frac{\frac{x}{x + 1}}{\frac{x \ln (x + 1) + x + \ln (x + 1)}{x + 1}}$

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 0} \frac{x}{x + 1} \cdot \frac{x + 1}{x \ln (x + 1) + x + \ln (x + 1)}$

Step 2.5

Multiply $\frac{x}{x + 1}$ by $\frac{x + 1}{x \ln (x + 1) + x + \ln (x + 1)}$ .

$lim_{x \to 0} \frac{x (x + 1)}{(x + 1) (x \ln (x + 1) + x + \ln (x + 1))}$

Step 2.6

Cancel the common factor of $x + 1$ .

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Step 2.6.1

Cancel the common factor.

$lim_{x \to 0} \frac{x (x + 1)}{(x + 1) (x \ln (x + 1) + x + \ln (x + 1))}$

Step 2.6.2

Rewrite the expression.

$lim_{x \to 0} \frac{x}{x \ln (x + 1) + x + \ln (x + 1)}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$\frac{lim_{x \to 0} x}{lim_{x \to 0} x \ln (x + 1) + x + \ln (x + 1)}$

Step 3.1.2

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

$\frac{0}{lim_{x \to 0} x \ln (x + 1) + x + \ln (x + 1)}$

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

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

$\frac{0}{lim_{x \to 0} x \ln (x + 1) + lim_{x \to 0} x + lim_{x \to 0} \ln (x + 1)}$

Step 3.1.3.2

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

$\frac{0}{lim_{x \to 0} x \cdot lim_{x \to 0} \ln (x + 1) + lim_{x \to 0} x + lim_{x \to 0} \ln (x + 1)}$

Step 3.1.3.3

Move the limit inside the logarithm.

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + lim_{x \to 0} \ln (x + 1)}$

Step 3.1.3.4

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + lim_{x \to 0} 1) + lim_{x \to 0} x + lim_{x \to 0} \ln (x + 1)}$

Step 3.1.3.5

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + lim_{x \to 0} \ln (x + 1)}$

Step 3.1.3.6

Move the limit inside the logarithm.

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + \ln (lim_{x \to 0} x + 1)}$

Step 3.1.3.7

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + \ln (lim_{x \to 0} x + lim_{x \to 0} 1)}$

Step 3.1.3.8

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

$\frac{0}{lim_{x \to 0} x \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + \ln (lim_{x \to 0} x + 1)}$

Step 3.1.3.9

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

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Step 3.1.3.9.1

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

$\frac{0}{0 \cdot \ln (lim_{x \to 0} x + 1) + lim_{x \to 0} x + \ln (lim_{x \to 0} x + 1)}$

Step 3.1.3.9.2

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

$\frac{0}{0 \cdot \ln (0 + 1) + lim_{x \to 0} x + \ln (lim_{x \to 0} x + 1)}$

Step 3.1.3.9.3

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

$\frac{0}{0 \cdot \ln (0 + 1) + 0 + \ln (lim_{x \to 0} x + 1)}$

Step 3.1.3.9.4

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

$\frac{0}{0 \cdot \ln (0 + 1) + 0 + \ln (0 + 1)}$

Step 3.1.3.10

Simplify the answer.

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Step 3.1.3.10.1

Simplify each term.

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Step 3.1.3.10.1.1

Add $0$ and $1$ .

$\frac{0}{0 \cdot \ln (1) + 0 + \ln (0 + 1)}$

Step 3.1.3.10.1.2

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

$\frac{0}{0 \cdot 0 + 0 + \ln (0 + 1)}$

Step 3.1.3.10.1.3

Multiply $0$ by $0$ .

$\frac{0}{0 + 0 + \ln (0 + 1)}$

Step 3.1.3.10.1.4

Add $0$ and $1$ .

$\frac{0}{0 + 0 + \ln (1)}$

Step 3.1.3.10.1.5

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

$\frac{0}{0 + 0 + 0}$

Step 3.1.3.10.2

Add $0$ and $0$ .

$\frac{0}{0 + 0}$

Step 3.1.3.10.3

Add $0$ and $0$ .

$\frac{0}{0}$

Step 3.1.3.10.4

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

Undefined

$\frac{0}{0}$

Step 3.1.3.11

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

Undefined

$\frac{0}{0}$

Step 3.1.4

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

Undefined

$\frac{0}{0}$

Step 3.2

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Evaluate $\frac{d}{d x} [x \ln (x + 1)]$ .

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Step 3.3.4.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (x + 1)$ .

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

Step 3.3.4.2

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 $g (x) = x + 1$ .

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Step 3.3.4.2.1

To apply the Chain Rule, set $u_{1}$ as $x + 1$ .

$lim_{x \to 0} \frac{1}{x (\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [x + 1]) + \ln (x + 1) \frac{d}{d x} [x] + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.2.2

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

$lim_{x \to 0} \frac{1}{x (\frac{1}{u_{1}} \frac{d}{d x} [x + 1]) + \ln (x + 1) \frac{d}{d x} [x] + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.2.3

Replace all occurrences of $u_{1}$ with $x + 1$ .

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

Step 3.3.4.3

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

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

Step 3.3.4.4

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

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

Step 3.3.4.5

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

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

Step 3.3.4.6

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

$lim_{x \to 0} \frac{1}{x (\frac{1}{x + 1} (1 + 0)) + \ln (x + 1) \cdot 1 + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.7

Add $1$ and $0$ .

$lim_{x \to 0} \frac{1}{x (\frac{1}{x + 1} \cdot 1) + \ln (x + 1) \cdot 1 + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.8

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

$lim_{x \to 0} \frac{1}{x \frac{1}{x + 1} + \ln (x + 1) \cdot 1 + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.9

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

$lim_{x \to 0} \frac{1}{\frac{x}{x + 1} + \ln (x + 1) \cdot 1 + \frac{d}{d x} [x] + \frac{d}{d x} [\ln (x + 1)]}$

Step 3.3.4.10

Multiply $\ln (x + 1)$ by $1$ .

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

Step 3.3.4.11

To write $\ln (x + 1)$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

Step 3.3.4.12

Combine the numerators over the common denominator.

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

Step 3.3.5

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

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

Step 3.3.6

Evaluate $\frac{d}{d x} [\ln (x + 1)]$ .

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Step 3.3.6.1

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

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Step 3.3.6.1.1

To apply the Chain Rule, set $u_{2}$ as $x + 1$ .

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + 1 + \frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [x + 1]}$

Step 3.3.6.1.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + 1 + \frac{1}{u_{2}} \frac{d}{d x} [x + 1]}$

Step 3.3.6.1.3

Replace all occurrences of $u_{2}$ with $x + 1$ .

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

Step 3.3.6.2

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

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

Step 3.3.6.3

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

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

Step 3.3.6.4

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

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + 1 + \frac{1}{x + 1} (1 + 0)}$

Step 3.3.6.5

Add $1$ and $0$ .

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + 1 + \frac{1}{x + 1} \cdot 1}$

Step 3.3.6.6

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

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + 1 + \frac{1}{x + 1}}$

Step 3.3.7

Simplify.

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Step 3.3.7.1

Combine terms.

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Step 3.3.7.1.1

Write $1$ as a fraction with a common denominator.

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1)}{x + 1} + \frac{x + 1}{x + 1} + \frac{1}{x + 1}}$

Step 3.3.7.1.2

Combine the numerators over the common denominator.

$lim_{x \to 0} \frac{1}{\frac{x + \ln (x + 1) (x + 1) + x + 1}{x + 1} + \frac{1}{x + 1}}$

Step 3.3.7.1.3

Add $x$ and $x$ .

$lim_{x \to 0} \frac{1}{\frac{\ln (x + 1) (x + 1) + 2 x + 1}{x + 1} + \frac{1}{x + 1}}$

Step 3.3.7.1.4

Combine the numerators over the common denominator.

$lim_{x \to 0} \frac{1}{\frac{\ln (x + 1) (x + 1) + 2 x + 1 + 1}{x + 1}}$

Step 3.3.7.1.5

Add $1$ and $1$ .

$lim_{x \to 0} \frac{1}{\frac{\ln (x + 1) (x + 1) + 2 x + 2}{x + 1}}$

Step 3.3.7.2

Simplify the numerator.

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Step 3.3.7.2.1

Apply the distributive property.

$lim_{x \to 0} \frac{1}{\frac{\ln (x + 1) x + \ln (x + 1) \cdot 1 + 2 x + 2}{x + 1}}$

Step 3.3.7.2.2

Multiply $\ln (x + 1)$ by $1$ .

$lim_{x \to 0} \frac{1}{\frac{\ln (x + 1) x + \ln (x + 1) + 2 x + 2}{x + 1}}$

Step 3.3.7.2.3

Rewrite $\ln (x + 1) x + \ln (x + 1) + 2 x + 2$ in a factored form.

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Step 3.3.7.2.3.1

Reorder terms.

$lim_{x \to 0} \frac{1}{\frac{x \ln (x + 1) + 2 x + \ln (x + 1) + 2}{x + 1}}$

Step 3.3.7.2.3.2

Factor out the greatest common factor from each group.

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Step 3.3.7.2.3.2.1

Group the first two terms and the last two terms.

$lim_{x \to 0} \frac{1}{\frac{(x \ln (x + 1) + 2 x) + \ln (x + 1) + 2}{x + 1}}$

Step 3.3.7.2.3.2.2

Factor out the greatest common factor (GCF) from each group.

$lim_{x \to 0} \frac{1}{\frac{x (\ln (x + 1) + 2) + 1 (\ln (x + 1) + 2)}{x + 1}}$

Step 3.3.7.2.3.3

Factor the polynomial by factoring out the greatest common factor, $\ln (x + 1) + 2$ .

$lim_{x \to 0} \frac{1}{\frac{(\ln (x + 1) + 2) (x + 1)}{x + 1}}$

Step 3.3.7.3

Cancel the common factor of $x + 1$ .

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Step 3.3.7.3.1

Cancel the common factor.

$lim_{x \to 0} \frac{1}{\frac{(\ln (x + 1) + 2) (x + 1)}{x + 1}}$

Step 3.3.7.3.2

Divide $\ln (x + 1) + 2$ by $1$ .

$lim_{x \to 0} \frac{1}{\ln (x + 1) + 2}$

Step 4

Evaluate the limit.

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Step 4.1

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

$\frac{lim_{x \to 0} 1}{lim_{x \to 0} \ln (x + 1) + 2}$

Step 4.2

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

$\frac{1}{lim_{x \to 0} \ln (x + 1) + 2}$

Step 4.3

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

$\frac{1}{lim_{x \to 0} \ln (x + 1) + lim_{x \to 0} 2}$

Step 4.4

Move the limit inside the logarithm.

$\frac{1}{\ln (lim_{x \to 0} x + 1) + lim_{x \to 0} 2}$

Step 4.5

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

$\frac{1}{\ln (lim_{x \to 0} x + lim_{x \to 0} 1) + lim_{x \to 0} 2}$

Step 4.6

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

$\frac{1}{\ln (lim_{x \to 0} x + 1) + lim_{x \to 0} 2}$

Step 4.7

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$\frac{1}{\ln (lim_{x \to 0} x + 1) + 2}$

Step 5

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

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

Step 6

Simplify the denominator.

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Step 6.1

Add $0$ and $1$ .

$\frac{1}{\ln (1) + 2}$

Step 6.2

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

$\frac{1}{0 + 2}$

Step 6.3

Add $0$ and $2$ .

$\frac{1}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{2}$

Decimal Form:

$0.5$