Calculus Examples

$lim_{x \to 9} \frac{1}{x - 9} - \frac{1}{\ln (x - 8)}$

Step 1

Combine terms.

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Step 1.1

To write $\frac{1}{x - 9}$ as a fraction with a common denominator, multiply by $\frac{\ln (x - 8)}{\ln (x - 8)}$ .

$lim_{x \to 9} \frac{1}{x - 9} \cdot \frac{\ln (x - 8)}{\ln (x - 8)} - \frac{1}{\ln (x - 8)}$

Step 1.2

To write $- \frac{1}{\ln (x - 8)}$ as a fraction with a common denominator, multiply by $\frac{x - 9}{x - 9}$ .

$lim_{x \to 9} \frac{1}{x - 9} \cdot \frac{\ln (x - 8)}{\ln (x - 8)} - \frac{1}{\ln (x - 8)} \cdot \frac{x - 9}{x - 9}$

Step 1.3

Write each expression with a common denominator of $(x - 9) \ln (x - 8)$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{1}{x - 9}$ by $\frac{\ln (x - 8)}{\ln (x - 8)}$ .

$lim_{x \to 9} \frac{\ln (x - 8)}{(x - 9) \ln (x - 8)} - \frac{1}{\ln (x - 8)} \cdot \frac{x - 9}{x - 9}$

Step 1.3.2

Multiply $\frac{1}{\ln (x - 8)}$ by $\frac{x - 9}{x - 9}$ .

$lim_{x \to 9} \frac{\ln (x - 8)}{(x - 9) \ln (x - 8)} - \frac{x - 9}{\ln (x - 8) (x - 9)}$

Step 1.3.3

Reorder the factors of $(x - 9) \ln (x - 8)$ .

$lim_{x \to 9} \frac{\ln (x - 8)}{\ln (x - 8) (x - 9)} - \frac{x - 9}{\ln (x - 8) (x - 9)}$

Step 1.4

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{\ln (x - 8) - (x - 9)}{\ln (x - 8) (x - 9)}$

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.

$\frac{lim_{x \to 9} \ln (x - 8) - (x - 9)}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2

Evaluate the limits by plugging in $9$ for all occurrences of $x$ .

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Step 2.1.2.1

Evaluate the limit of $\ln (x - 8) - (x - 9)$ by plugging in $9$ for $x$ .

$\frac{\ln (9 - 8) - (9 - 9)}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2.2

Simplify each term.

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Step 2.1.2.2.1

Subtract $8$ from $9$ .

$\frac{\ln (1) - (9 - 9)}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2.2.2

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

$\frac{0 - (9 - 9)}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2.2.3

Subtract $9$ from $9$ .

$\frac{0 - 0}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2.2.4

Multiply $- 1$ by $0$ .

$\frac{0 + 0}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

Step 2.1.2.3

Add $0$ and $0$ .

$\frac{0}{lim_{x \to 9} \ln (x - 8) (x - 9)}$

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 $9$ .

$\frac{0}{lim_{x \to 9} \ln (x - 8) \cdot lim_{x \to 9} x - 9}$

Step 2.1.3.2

Move the limit inside the logarithm.

$\frac{0}{\ln (lim_{x \to 9} x - 8) \cdot lim_{x \to 9} x - 9}$

Step 2.1.3.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{\ln (lim_{x \to 9} x - lim_{x \to 9} 8) \cdot lim_{x \to 9} x - 9}$

Step 2.1.3.4

Evaluate the limit of $8$ which is constant as $x$ approaches $9$ .

$\frac{0}{\ln (lim_{x \to 9} x - 1 \cdot 8) \cdot lim_{x \to 9} x - 9}$

Step 2.1.3.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{\ln (lim_{x \to 9} x - 1 \cdot 8) \cdot (lim_{x \to 9} x - lim_{x \to 9} 9)}$

Step 2.1.3.6

Evaluate the limit of $9$ which is constant as $x$ approaches $9$ .

$\frac{0}{\ln (lim_{x \to 9} x - 1 \cdot 8) \cdot (lim_{x \to 9} x - 1 \cdot 9)}$

Step 2.1.3.7

Evaluate the limits by plugging in $9$ for all occurrences of $x$ .

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Step 2.1.3.7.1

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{\ln (9 - 1 \cdot 8) \cdot (lim_{x \to 9} x - 1 \cdot 9)}$

Step 2.1.3.7.2

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{\ln (9 - 1 \cdot 8) \cdot (9 - 1 \cdot 9)}$

Step 2.1.3.8

Simplify the answer.

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Step 2.1.3.8.1

Multiply $- 1$ by $8$ .

$\frac{0}{\ln (9 - 8) \cdot (9 - 1 \cdot 9)}$

Step 2.1.3.8.2

Subtract $8$ from $9$ .

$\frac{0}{\ln (1) \cdot (9 - 1 \cdot 9)}$

Step 2.1.3.8.3

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

$\frac{0}{0 \cdot (9 - 1 \cdot 9)}$

Step 2.1.3.8.4

Multiply $- 1$ by $9$ .

$\frac{0}{0 \cdot (9 - 9)}$

Step 2.1.3.8.5

Subtract $9$ from $9$ .

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

Step 2.1.3.8.6

Multiply $0$ by $0$ .

$\frac{0}{0}$

Step 2.1.3.8.7

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

Undefined

$\frac{0}{0}$

Step 2.1.3.9

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.

$lim_{x \to 9} \frac{\ln (x - 8) - (x - 9)}{\ln (x - 8) (x - 9)} = lim_{x \to 9} \frac{\frac{d}{d x} [\ln (x - 8) - (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to 9} \frac{\frac{d}{d x} [\ln (x - 8) - (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.2

By the Sum Rule, the derivative of $\ln (x - 8) - (x - 9)$ with respect to $x$ is $\frac{d}{d x} [\ln (x - 8)] + \frac{d}{d x} [- (x - 9)]$ .

$lim_{x \to 9} \frac{\frac{d}{d x} [\ln (x - 8)] + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3

Evaluate $\frac{d}{d x} [\ln (x - 8)]$ .

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Step 2.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x - 8$ .

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Step 2.3.3.1.1

To apply the Chain Rule, set $u$ as $x - 8$ .

$lim_{x \to 9} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x - 8] + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.1.2

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

$lim_{x \to 9} \frac{\frac{1}{u} \frac{d}{d x} [x - 8] + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.1.3

Replace all occurrences of $u$ with $x - 8$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} \frac{d}{d x} [x - 8] + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.2

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} (\frac{d}{d x} [x] + \frac{d}{d x} [- 8]) + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.3

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

$lim_{x \to 9} \frac{\frac{1}{x - 8} (1 + \frac{d}{d x} [- 8]) + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.4

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} (1 + 0) + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.5

Add $1$ and $0$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} \cdot 1 + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.3.6

Multiply $\frac{1}{x - 8}$ by $1$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} + \frac{d}{d x} [- (x - 9)]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4

Evaluate $\frac{d}{d x} [- (x - 9)]$ .

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Step 2.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x - 9)$ with respect to $x$ is $- \frac{d}{d x} [x - 9]$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - \frac{d}{d x} [x - 9]}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4.2

By the Sum Rule, the derivative of $x - 9$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 9]$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - (\frac{d}{d x} [x] + \frac{d}{d x} [- 9])}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4.3

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

$lim_{x \to 9} \frac{\frac{1}{x - 8} - (1 + \frac{d}{d x} [- 9])}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4.4

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - (1 + 0)}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4.5

Add $1$ and $0$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - 1 \cdot 1}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.4.6

Multiply $- 1$ by $1$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - 1}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.5

Combine terms.

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Step 2.3.5.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x - 8}{x - 8}$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} - 1 \cdot \frac{x - 8}{x - 8}}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.5.2

Combine $- 1$ and $\frac{x - 8}{x - 8}$ .

$lim_{x \to 9} \frac{\frac{1}{x - 8} + \frac{- (x - 8)}{x - 8}}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

Step 2.3.5.3

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\frac{d}{d x} [\ln (x - 8) (x - 9)]}$

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 $f (x) = \ln (x - 8)$ and $g (x) = x - 9$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) \frac{d}{d x} [x - 9] + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.7

By the Sum Rule, the derivative of $x - 9$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 9]$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) (\frac{d}{d x} [x] + \frac{d}{d x} [- 9]) + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.8

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

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) (1 + \frac{d}{d x} [- 9]) + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.9

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) (1 + 0) + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.10

Add $1$ and $0$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) \cdot 1 + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.11

Multiply $\ln (x - 8)$ by $1$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{d}{d x} [\ln (x - 8)]}$

Step 2.3.12

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 $g (x) = x - 8$ .

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Step 2.3.12.1

To apply the Chain Rule, set $u$ as $x - 8$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x - 8])}$

Step 2.3.12.2

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

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) (\frac{1}{u} \frac{d}{d x} [x - 8])}$

Step 2.3.12.3

Replace all occurrences of $u$ with $x - 8$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) (\frac{1}{x - 8} \frac{d}{d x} [x - 8])}$

Step 2.3.13

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{1}{x - 8} (\frac{d}{d x} [x] + \frac{d}{d x} [- 8])}$

Step 2.3.14

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

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{1}{x - 8} (1 + \frac{d}{d x} [- 8])}$

Step 2.3.15

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{1}{x - 8} (1 + 0)}$

Step 2.3.16

Add $1$ and $0$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{1}{x - 8} \cdot 1}$

Step 2.3.17

Multiply $x - 9$ by $1$ .

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{\ln (x - 8) + (x - 9) \frac{1}{x - 8}}$

Step 2.3.18

Reorder terms.

$lim_{x \to 9} \frac{\frac{1 - (x - 8)}{x - 8}}{(x - 9) \frac{1}{x - 8} + \ln (x - 8)}$

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 9} \frac{1 - (x - 8)}{x - 8} \cdot \frac{1}{(x - 9) \frac{1}{x - 8} + \ln (x - 8)}$

Step 2.5

Multiply $\frac{1 - (x - 8)}{x - 8}$ by $\frac{1}{(x - 9) \frac{1}{x - 8} + \ln (x - 8)}$ .

$lim_{x \to 9} \frac{1 - (x - 8)}{(x - 8) ((x - 9) \frac{1}{x - 8} + \ln (x - 8))}$

Step 2.6

Multiply $x - 9$ by $\frac{1}{x - 8}$ .

$lim_{x \to 9} \frac{1 - (x - 8)}{(x - 8) (\frac{x - 9}{x - 8} + \ln (x - 8))}$

Step 2.7

Combine terms.

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Step 2.7.1

To write $\ln (x - 8)$ as a fraction with a common denominator, multiply by $\frac{x - 8}{x - 8}$ .

$lim_{x \to 9} \frac{1 - (x - 8)}{(x - 8) (\frac{x - 9}{x - 8} + \frac{\ln (x - 8) (x - 8)}{x - 8})}$

Step 2.7.2

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{1 - (x - 8)}{(x - 8) \frac{x - 9 + \ln (x - 8) (x - 8)}{x - 8}}$

Step 3

Simplify the limit argument.

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Step 3.1

Multiply $x - 8$ by $\frac{x - 9 + \ln (x - 8) (x - 8)}{x - 8}$ .

$lim_{x \to 9} \frac{1 - (x - 8)}{\frac{(x - 8) (x - 9 + \ln (x - 8) (x - 8))}{x - 8}}$

Step 3.2

Cancel the common factor of $x - 8$ .

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Step 3.2.1

Cancel the common factor.

$lim_{x \to 9} \frac{1 - (x - 8)}{\frac{(x - 8) (x - 9 + \ln (x - 8) (x - 8))}{x - 8}}$

Step 3.2.2

Divide $x - 9 + \ln (x - 8) (x - 8)$ by $1$ .

$lim_{x \to 9} \frac{1 - (x - 8)}{x - 9 + \ln (x - 8) (x - 8)}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

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

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Step 4.1.1

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

$\frac{lim_{x \to 9} 1 - (x - 8)}{lim_{x \to 9} x - 9 + \ln (x - 8) (x - 8)}$

Step 4.1.2

Evaluate the limits by plugging in $9$ for all occurrences of $x$ .

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Step 4.1.2.1

Evaluate the limit of $1 - (x - 8)$ by plugging in $9$ for $x$ .

$\frac{1 - (9 - 8)}{lim_{x \to 9} x - 9 + \ln (x - 8) (x - 8)}$

Step 4.1.2.2

Simplify each term.

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Step 4.1.2.2.1

Subtract $8$ from $9$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to 9} x - 9 + \ln (x - 8) (x - 8)}$

Step 4.1.2.2.2

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 9} x - 9 + \ln (x - 8) (x - 8)}$

Step 4.1.2.3

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 9} x - 9 + \ln (x - 8) (x - 8)}$

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - lim_{x \to 9} 9 + lim_{x \to 9} \ln (x - 8) (x - 8)}$

Step 4.1.3.2

Evaluate the limit of $9$ which is constant as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + lim_{x \to 9} \ln (x - 8) (x - 8)}$

Step 4.1.3.3

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + lim_{x \to 9} \ln (x - 8) \cdot lim_{x \to 9} x - 8}$

Step 4.1.3.4

Move the limit inside the logarithm.

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + \ln (lim_{x \to 9} x - 8) \cdot lim_{x \to 9} x - 8}$

Step 4.1.3.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + \ln (lim_{x \to 9} x - lim_{x \to 9} 8) \cdot lim_{x \to 9} x - 8}$

Step 4.1.3.6

Evaluate the limit of $8$ which is constant as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + \ln (lim_{x \to 9} x - 1 \cdot 8) \cdot lim_{x \to 9} x - 8}$

Step 4.1.3.7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + \ln (lim_{x \to 9} x - 1 \cdot 8) \cdot (lim_{x \to 9} x - lim_{x \to 9} 8)}$

Step 4.1.3.8

Evaluate the limit of $8$ which is constant as $x$ approaches $9$ .

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9 + \ln (lim_{x \to 9} x - 1 \cdot 8) \cdot (lim_{x \to 9} x - 1 \cdot 8)}$

Step 4.1.3.9

Evaluate the limits by plugging in $9$ for all occurrences of $x$ .

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Step 4.1.3.9.1

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{9 - 1 \cdot 9 + \ln (lim_{x \to 9} x - 1 \cdot 8) \cdot (lim_{x \to 9} x - 1 \cdot 8)}$

Step 4.1.3.9.2

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{9 - 1 \cdot 9 + \ln (9 - 1 \cdot 8) \cdot (lim_{x \to 9} x - 1 \cdot 8)}$

Step 4.1.3.9.3

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{0}{9 - 1 \cdot 9 + \ln (9 - 1 \cdot 8) \cdot (9 - 1 \cdot 8)}$

Step 4.1.3.10

Simplify the answer.

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Step 4.1.3.10.1

Simplify each term.

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Step 4.1.3.10.1.1

Multiply $- 1$ by $9$ .

$\frac{0}{9 - 9 + \ln (9 - 1 \cdot 8) \cdot (9 - 1 \cdot 8)}$

Step 4.1.3.10.1.2

Multiply $- 1$ by $8$ .

$\frac{0}{9 - 9 + \ln (9 - 8) \cdot (9 - 1 \cdot 8)}$

Step 4.1.3.10.1.3

Subtract $8$ from $9$ .

$\frac{0}{9 - 9 + \ln (1) \cdot (9 - 1 \cdot 8)}$

Step 4.1.3.10.1.4

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

$\frac{0}{9 - 9 + 0 \cdot (9 - 1 \cdot 8)}$

Step 4.1.3.10.1.5

Multiply $- 1$ by $8$ .

$\frac{0}{9 - 9 + 0 \cdot (9 - 8)}$

Step 4.1.3.10.1.6

Subtract $8$ from $9$ .

$\frac{0}{9 - 9 + 0 \cdot 1}$

Step 4.1.3.10.1.7

Multiply $0$ by $1$ .

$\frac{0}{9 - 9 + 0}$

Step 4.1.3.10.2

Subtract $9$ from $9$ .

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

Step 4.1.3.10.3

Add $0$ and $0$ .

$\frac{0}{0}$

Step 4.1.3.10.4

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

Undefined

$\frac{0}{0}$

Step 4.1.3.11

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

Undefined

$\frac{0}{0}$

Step 4.1.4

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

Undefined

$\frac{0}{0}$

Step 4.2

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.

$lim_{x \to 9} \frac{1 - (x - 8)}{x - 9 + \ln (x - 8) (x - 8)} = lim_{x \to 9} \frac{\frac{d}{d x} [1 - (x - 8)]}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$lim_{x \to 9} \frac{\frac{d}{d x} [1 - (x - 8)]}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.2

By the Sum Rule, the derivative of $1 - (x - 8)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- (x - 8)]$ .

$lim_{x \to 9} \frac{\frac{d}{d x} [1] + \frac{d}{d x} [- (x - 8)]}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.3

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

$lim_{x \to 9} \frac{0 + \frac{d}{d x} [- (x - 8)]}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4

Evaluate $\frac{d}{d x} [- (x - 8)]$ .

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Step 4.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x - 8)$ with respect to $x$ is $- \frac{d}{d x} [x - 8]$ .

$lim_{x \to 9} \frac{0 - \frac{d}{d x} [x - 8]}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4.2

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 9} \frac{0 - (\frac{d}{d x} [x] + \frac{d}{d x} [- 8])}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4.3

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

$lim_{x \to 9} \frac{0 - (1 + \frac{d}{d x} [- 8])}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4.4

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{0 - (1 + 0)}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4.5

Add $1$ and $0$ .

$lim_{x \to 9} \frac{0 - 1 \cdot 1}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.4.6

Multiply $- 1$ by $1$ .

$lim_{x \to 9} \frac{0 - 1}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.5

Subtract $1$ from $0$ .

$lim_{x \to 9} \frac{- 1}{\frac{d}{d x} [x - 9 + \ln (x - 8) (x - 8)]}$

Step 4.3.6

By the Sum Rule, the derivative of $x - 9 + \ln (x - 8) (x - 8)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 9] + \frac{d}{d x} [\ln (x - 8) (x - 8)]$ .

$lim_{x \to 9} \frac{- 1}{\frac{d}{d x} [x] + \frac{d}{d x} [- 9] + \frac{d}{d x} [\ln (x - 8) (x - 8)]}$

Step 4.3.7

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

$lim_{x \to 9} \frac{- 1}{1 + \frac{d}{d x} [- 9] + \frac{d}{d x} [\ln (x - 8) (x - 8)]}$

Step 4.3.8

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \frac{d}{d x} [\ln (x - 8) (x - 8)]}$

Step 4.3.9

Evaluate $\frac{d}{d x} [\ln (x - 8) (x - 8)]$ .

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Step 4.3.9.1

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 $f (x) = \ln (x - 8)$ and $g (x) = x - 8$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) \frac{d}{d x} [x - 8] + (x - 8) \frac{d}{d x} [\ln (x - 8)]}$

Step 4.3.9.2

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (\frac{d}{d x} [x] + \frac{d}{d x} [- 8]) + (x - 8) \frac{d}{d x} [\ln (x - 8)]}$

Step 4.3.9.3

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

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + \frac{d}{d x} [- 8]) + (x - 8) \frac{d}{d x} [\ln (x - 8)]}$

Step 4.3.9.4

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) \frac{d}{d x} [\ln (x - 8)]}$

Step 4.3.9.5

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 $g (x) = x - 8$ .

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Step 4.3.9.5.1

To apply the Chain Rule, set $u$ as $x - 8$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x - 8])}$

Step 4.3.9.5.2

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

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{1}{u} \frac{d}{d x} [x - 8])}$

Step 4.3.9.5.3

Replace all occurrences of $u$ with $x - 8$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{1}{x - 8} \frac{d}{d x} [x - 8])}$

Step 4.3.9.6

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{1}{x - 8} (\frac{d}{d x} [x] + \frac{d}{d x} [- 8]))}$

Step 4.3.9.7

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

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{1}{x - 8} (1 + \frac{d}{d x} [- 8]))}$

Step 4.3.9.8

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) (1 + 0) + (x - 8) (\frac{1}{x - 8} (1 + 0))}$

Step 4.3.9.9

Add $1$ and $0$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) \cdot 1 + (x - 8) (\frac{1}{x - 8} (1 + 0))}$

Step 4.3.9.10

Multiply $\ln (x - 8)$ by $1$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) + (x - 8) (\frac{1}{x - 8} (1 + 0))}$

Step 4.3.9.11

Add $1$ and $0$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) + (x - 8) (\frac{1}{x - 8} \cdot 1)}$

Step 4.3.9.12

Multiply $\frac{1}{x - 8}$ by $1$ .

$lim_{x \to 9} \frac{- 1}{1 + 0 + \ln (x - 8) + (x - 8) \frac{1}{x - 8}}$

Step 4.3.10

Simplify.

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Step 4.3.10.1

Add $1$ and $0$ .

$lim_{x \to 9} \frac{- 1}{1 + \ln (x - 8) + (x - 8) \frac{1}{x - 8}}$

Step 4.3.10.2

Reorder terms.

$lim_{x \to 9} \frac{- 1}{(x - 8) \frac{1}{x - 8} + 1 + \ln (x - 8)}$

Step 4.3.10.3

Cancel the common factor of $x - 8$ .

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Step 4.3.10.3.1

Cancel the common factor.

$lim_{x \to 9} \frac{- 1}{(x - 8) \frac{1}{x - 8} + 1 + \ln (x - 8)}$

Step 4.3.10.3.2

Rewrite the expression.

$lim_{x \to 9} \frac{- 1}{1 + 1 + \ln (x - 8)}$

Step 4.3.10.4

Add $1$ and $1$ .

$lim_{x \to 9} \frac{- 1}{2 + \ln (x - 8)}$

Step 5

Evaluate the limit.

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Step 5.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $9$ .

$\frac{lim_{x \to 9} - 1}{lim_{x \to 9} 2 + \ln (x - 8)}$

Step 5.2

Evaluate the limit of $- 1$ which is constant as $x$ approaches $9$ .

$\frac{- 1}{lim_{x \to 9} 2 + \ln (x - 8)}$

Step 5.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{- 1}{lim_{x \to 9} 2 + lim_{x \to 9} \ln (x - 8)}$

Step 5.4

Evaluate the limit of $2$ which is constant as $x$ approaches $9$ .

$\frac{- 1}{2 + lim_{x \to 9} \ln (x - 8)}$

Step 5.5

Move the limit inside the logarithm.

$\frac{- 1}{2 + \ln (lim_{x \to 9} x - 8)}$

Step 5.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $9$ .

$\frac{- 1}{2 + \ln (lim_{x \to 9} x - lim_{x \to 9} 8)}$

Step 5.7

Evaluate the limit of $8$ which is constant as $x$ approaches $9$ .

$\frac{- 1}{2 + \ln (lim_{x \to 9} x - 1 \cdot 8)}$

Step 6

Evaluate the limit of $x$ by plugging in $9$ for $x$ .

$\frac{- 1}{2 + \ln (9 - 1 \cdot 8)}$

Step 7

Simplify the answer.

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Step 7.1

Simplify the denominator.

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Step 7.1.1

Multiply $- 1$ by $8$ .

$\frac{- 1}{2 + \ln (9 - 8)}$

Step 7.1.2

Subtract $8$ from $9$ .

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

Step 7.1.3

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

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

Step 7.1.4

Add $2$ and $0$ .

$\frac{- 1}{2}$

Step 7.2

Move the negative in front of the fraction.

$- \frac{1}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{2}$

Decimal Form:

$- 0.5$