Calculus Examples

$lim_{x \to 2} \frac{\ln (x + 3) - \ln (5)}{x - 2}$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$lim_{x \to 2} \frac{\ln (\frac{x + 3}{5})}{x - 2}$

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.

$\frac{lim_{x \to 2} \ln (\frac{x + 3}{5})}{lim_{x \to 2} x - 2}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

Move the limit inside the logarithm.

$\frac{\ln (lim_{x \to 2} \frac{x + 3}{5})}{lim_{x \to 2} x - 2}$

Step 2.1.2.1.2

Move the term $\frac{1}{5}$ outside of the limit because it is constant with respect to $x$ .

$\frac{\ln (\frac{1}{5} lim_{x \to 2} x + 3)}{lim_{x \to 2} x - 2}$

Step 2.1.2.1.3

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

$\frac{\ln (\frac{1}{5} (lim_{x \to 2} x + lim_{x \to 2} 3))}{lim_{x \to 2} x - 2}$

Step 2.1.2.1.4

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

$\frac{\ln (\frac{1}{5} (lim_{x \to 2} x + 3))}{lim_{x \to 2} x - 2}$

Step 2.1.2.2

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

$\frac{\ln (\frac{1}{5} (2 + 3))}{lim_{x \to 2} x - 2}$

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Add $2$ and $3$ .

$\frac{\ln (\frac{1}{5} \cdot 5)}{lim_{x \to 2} x - 2}$

Step 2.1.2.3.2

Cancel the common factor of $5$ .

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Step 2.1.2.3.2.1

Cancel the common factor.

$\frac{\ln (\frac{1}{5} \cdot 5)}{lim_{x \to 2} x - 2}$

Step 2.1.2.3.2.2

Rewrite the expression.

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

Step 2.1.2.3.3

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

$\frac{0}{lim_{x \to 2} x - 2}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Evaluate the limit.

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Step 2.1.3.1.1

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

$\frac{0}{lim_{x \to 2} x - lim_{x \to 2} 2}$

Step 2.1.3.1.2

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

Simplify the answer.

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Step 2.1.3.3.1

Multiply $- 1$ by $2$ .

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

Step 2.1.3.3.2

Subtract $2$ from $2$ .

$\frac{0}{0}$

Step 2.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 2.1.3.4

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.

$lim_{x \to 2} \frac{\ln (\frac{x + 3}{5})}{x - 2} = lim_{x \to 2} \frac{\frac{d}{d x} [\ln (\frac{x + 3}{5})]}{\frac{d}{d x} [x - 2]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to 2} \frac{\frac{d}{d x} [\ln (\frac{x + 3}{5})]}{\frac{d}{d x} [x - 2]}$

Step 2.3.2

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 $g (x) = \frac{x + 3}{5}$ .

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $\frac{x + 3}{5}$ .

$lim_{x \to 2} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{x + 3}{5}]}{\frac{d}{d x} [x - 2]}$

Step 2.3.2.2

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

$lim_{x \to 2} \frac{\frac{1}{u} \frac{d}{d x} [\frac{x + 3}{5}]}{\frac{d}{d x} [x - 2]}$

Step 2.3.2.3

Replace all occurrences of $u$ with $\frac{x + 3}{5}$ .

$lim_{x \to 2} \frac{\frac{1}{\frac{x + 3}{5}} \frac{d}{d x} [\frac{x + 3}{5}]}{\frac{d}{d x} [x - 2]}$

Step 2.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{x + 3}{5}$ .

$lim_{x \to 2} \frac{1 \frac{5}{x + 3} \frac{d}{d x} [\frac{x + 3}{5}]}{\frac{d}{d x} [x - 2]}$

Step 2.3.4

Multiply $\frac{5}{x + 3}$ by $1$ .

$lim_{x \to 2} \frac{\frac{5}{x + 3} \frac{d}{d x} [\frac{x + 3}{5}]}{\frac{d}{d x} [x - 2]}$

Step 2.3.5

Since $\frac{1}{5}$ is constant with respect to $x$ , the derivative of $\frac{x + 3}{5}$ with respect to $x$ is $\frac{1}{5} \frac{d}{d x} [x + 3]$ .

$lim_{x \to 2} \frac{\frac{5}{x + 3} (\frac{1}{5} \frac{d}{d x} [x + 3])}{\frac{d}{d x} [x - 2]}$

Step 2.3.6

Multiply $\frac{1}{5}$ by $\frac{5}{x + 3}$ .

$lim_{x \to 2} \frac{\frac{5}{5 (x + 3)} \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x - 2]}$

Step 2.3.7

Cancel the common factor of $5$ .

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Step 2.3.7.1

Cancel the common factor.

$lim_{x \to 2} \frac{\frac{5}{5 (x + 3)} \frac{d}{d x} [x + 3]}{\frac{d}{d x} [x - 2]}$

Step 2.3.7.2

Rewrite the expression.

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

Step 2.3.8

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

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

Step 2.3.9

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

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

Step 2.3.10

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

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

Step 2.3.11

Add $1$ and $0$ .

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

Step 2.3.12

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

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

Step 2.3.13

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

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

Step 2.3.14

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

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

Step 2.3.15

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

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

Step 2.3.16

Add $1$ and $0$ .

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

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.5

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

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

Step 3

Evaluate the limit.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

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

Step 4

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

$\frac{1}{2 + 3}$

Step 5

Add $2$ and $3$ .

$\frac{1}{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{5}$

Decimal Form:

$0.2$