Calculus Examples

$lim_{x \to 4} \frac{\sqrt{x^{2} + 1} - \sqrt{x + 13}}{x - 4}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{x \to 4} \sqrt{x^{2} + 1} - \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{lim_{x \to 4} \sqrt{x^{2} + 1} - lim_{x \to 4} \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.2

Move the limit under the radical sign.

$\frac{\sqrt{lim_{x \to 4} x^{2} + 1} - lim_{x \to 4} \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{\sqrt{lim_{x \to 4} x^{2} + lim_{x \to 4} 1} - lim_{x \to 4} \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.4

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{\sqrt{{(lim_{x \to 4} x)}^{2} + lim_{x \to 4} 1} - lim_{x \to 4} \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.5

Evaluate the limit of $1$ which is constant as $x$ approaches $4$ .

$\frac{\sqrt{{(lim_{x \to 4} x)}^{2} + 1} - lim_{x \to 4} \sqrt{x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.6

Move the limit under the radical sign.

$\frac{\sqrt{{(lim_{x \to 4} x)}^{2} + 1} - \sqrt{lim_{x \to 4} x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{\sqrt{{(lim_{x \to 4} x)}^{2} + 1} - \sqrt{lim_{x \to 4} x + lim_{x \to 4} 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.8

Evaluate the limit of $13$ which is constant as $x$ approaches $4$ .

$\frac{\sqrt{{(lim_{x \to 4} x)}^{2} + 1} - \sqrt{lim_{x \to 4} x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.9

Evaluate the limits by plugging in $4$ for all occurrences of $x$ .

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Step 1.1.2.9.1

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{\sqrt{4^{2} + 1} - \sqrt{lim_{x \to 4} x + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.9.2

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{\sqrt{4^{2} + 1} - \sqrt{4 + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.10

Simplify the answer.

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Step 1.1.2.10.1

Simplify each term.

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Step 1.1.2.10.1.1

Raise $4$ to the power of $2$ .

$\frac{\sqrt{16 + 1} - \sqrt{4 + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.10.1.2

Add $16$ and $1$ .

$\frac{\sqrt{17} - \sqrt{4 + 13}}{lim_{x \to 4} x - 4}$

Step 1.1.2.10.1.3

Add $4$ and $13$ .

$\frac{\sqrt{17} - \sqrt{17}}{lim_{x \to 4} x - 4}$

Step 1.1.2.10.2

Subtract $\sqrt{17}$ from $\sqrt{17}$ .

$\frac{0}{lim_{x \to 4} x - 4}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{0}{lim_{x \to 4} x - lim_{x \to 4} 4}$

Step 1.1.3.1.2

Evaluate the limit of $4$ which is constant as $x$ approaches $4$ .

$\frac{0}{lim_{x \to 4} x - 1 \cdot 4}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{0}{4 - 1 \cdot 4}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $- 1$ by $4$ .

$\frac{0}{4 - 4}$

Step 1.1.3.3.2

Subtract $4$ from $4$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 4} \frac{\sqrt{x^{2} + 1} - \sqrt{x + 13}}{x - 4} = lim_{x \to 4} \frac{\frac{d}{d x} [\sqrt{x^{2} + 1} - \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 4} \frac{\frac{d}{d x} [\sqrt{x^{2} + 1} - \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.2

By the Sum Rule, the derivative of $\sqrt{x^{2} + 1} - \sqrt{x + 13}$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x^{2} + 1}] + \frac{d}{d x} [- \sqrt{x + 13}]$ .

$lim_{x \to 4} \frac{\frac{d}{d x} [\sqrt{x^{2} + 1}] + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [\sqrt{x^{2} + 1}]$ .

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Step 1.3.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 1}$ as ${(x^{2} + 1)}^{\frac{1}{2}}$ .

$lim_{x \to 4} \frac{\frac{d}{d x} [{(x^{2} + 1)}^{\frac{1}{2}}] + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{2} + 1$ .

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Step 1.3.3.2.1

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

$lim_{x \to 4} \frac{\frac{d}{d u_{1}} [{u_{1}}^{\frac{1}{2}}] \frac{d}{d x} [x^{2} + 1] + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = \frac{1}{2}$ .

$lim_{x \to 4} \frac{\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + 1] + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.2.3

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

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + 1] + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.3

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} (2 x + \frac{d}{d x} [1]) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.5

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

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.7

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

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.8

Combine the numerators over the common denominator.

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1 - 1 \cdot 2}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.9

Simplify the numerator.

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Step 1.3.3.9.1

Multiply $- 1$ by $2$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{1 - 2}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.9.2

Subtract $2$ from $1$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{\frac{- 1}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.10

Move the negative in front of the fraction.

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{- \frac{1}{2}} (2 x + 0) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.11

Add $2 x$ and $0$ .

$lim_{x \to 4} \frac{\frac{1}{2} {(x^{2} + 1)}^{- \frac{1}{2}} (2 x) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.12

Combine $\frac{1}{2}$ and ${(x^{2} + 1)}^{- \frac{1}{2}}$ .

$lim_{x \to 4} \frac{\frac{{(x^{2} + 1)}^{- \frac{1}{2}}}{2} (2 x) + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.13

Combine $2$ and $\frac{{(x^{2} + 1)}^{- \frac{1}{2}}}{2}$ .

$lim_{x \to 4} \frac{\frac{2 {(x^{2} + 1)}^{- \frac{1}{2}}}{2} x + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.14

Combine $\frac{2 {(x^{2} + 1)}^{- \frac{1}{2}}}{2}$ and $x$ .

$lim_{x \to 4} \frac{\frac{2 {(x^{2} + 1)}^{- \frac{1}{2}} x}{2} + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.15

Move ${(x^{2} + 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 4} \frac{\frac{2 x}{2 {(x^{2} + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.16

Cancel the common factor.

$lim_{x \to 4} \frac{\frac{2 x}{2 {(x^{2} + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.3.17

Rewrite the expression.

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [- \sqrt{x + 13}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- \sqrt{x + 13}]$ .

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Step 1.3.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 13}$ as ${(x + 13)}^{\frac{1}{2}}$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [- {(x + 13)}^{\frac{1}{2}}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- {(x + 13)}^{\frac{1}{2}}$ with respect to $x$ is $- \frac{d}{d x} [{(x + 13)}^{\frac{1}{2}}]$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{d}{d x} [{(x + 13)}^{\frac{1}{2}}]}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 13$ .

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Step 1.3.4.3.1

To apply the Chain Rule, set $u_{2}$ as $x + 13$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d x} [x + 13])}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = \frac{1}{2}$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 13])}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.3.3

Replace all occurrences of $u_{2}$ with $x + 13$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 13])}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.4

By the Sum Rule, the derivative of $x + 13$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [13]$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [13]))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.5

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

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [13]))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.6

Since $13$ is constant with respect to $x$ , the derivative of $13$ with respect to $x$ is $0$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} - 1} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.8

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

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.9

Combine the numerators over the common denominator.

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1 - 1 \cdot 2}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.10

Simplify the numerator.

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Step 1.3.4.10.1

Multiply $- 1$ by $2$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{1 - 2}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.10.2

Subtract $2$ from $1$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{\frac{- 1}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.11

Move the negative in front of the fraction.

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{- \frac{1}{2}} (1 + 0))}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.12

Add $1$ and $0$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{1}{2} {(x + 13)}^{- \frac{1}{2}} \cdot 1)}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.13

Combine $\frac{1}{2}$ and ${(x + 13)}^{- \frac{1}{2}}$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - (\frac{{(x + 13)}^{- \frac{1}{2}}}{2} \cdot 1)}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.14

Multiply $\frac{{(x + 13)}^{- \frac{1}{2}}}{2}$ by $1$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{{(x + 13)}^{- \frac{1}{2}}}{2}}{\frac{d}{d x} [x - 4]}$

Step 1.3.4.15

Move ${(x + 13)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{\frac{d}{d x} [x - 4]}$

Step 1.3.5

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{\frac{d}{d x} [x] + \frac{d}{d x} [- 4]}$

Step 1.3.6

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

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{1 + \frac{d}{d x} [- 4]}$

Step 1.3.7

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{1 + 0}$

Step 1.3.8

Add $1$ and $0$ .

$lim_{x \to 4} \frac{\frac{x}{{(x^{2} + 1)}^{\frac{1}{2}}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{1}$

Step 1.4

Convert fractional exponents to radicals.

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Step 1.4.1

Rewrite ${(x^{2} + 1)}^{\frac{1}{2}}$ as $\sqrt{x^{2} + 1}$ .

$lim_{x \to 4} \frac{\frac{x}{\sqrt{x^{2} + 1}} - \frac{1}{2 {(x + 13)}^{\frac{1}{2}}}}{1}$

Step 1.4.2

Rewrite ${(x + 13)}^{\frac{1}{2}}$ as $\sqrt{x + 13}$ .

$lim_{x \to 4} \frac{\frac{x}{\sqrt{x^{2} + 1}} - \frac{1}{2 \sqrt{x + 13}}}{1}$

Step 1.5

Combine terms.

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Step 1.5.1

To write $\frac{x}{\sqrt{x^{2} + 1}}$ as a fraction with a common denominator, multiply by $\frac{2 \sqrt{x + 13}}{2 \sqrt{x + 13}}$ .

$lim_{x \to 4} \frac{\frac{x}{\sqrt{x^{2} + 1}} \cdot \frac{2 \sqrt{x + 13}}{2 \sqrt{x + 13}} - \frac{1}{2 \sqrt{x + 13}}}{1}$

Step 1.5.2

To write $- \frac{1}{2 \sqrt{x + 13}}$ as a fraction with a common denominator, multiply by $\frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}$ .

$lim_{x \to 4} \frac{\frac{x}{\sqrt{x^{2} + 1}} \cdot \frac{2 \sqrt{x + 13}}{2 \sqrt{x + 13}} - \frac{1}{2 \sqrt{x + 13}} \cdot \frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}}{1}$

Step 1.5.3

Write each expression with a common denominator of $\sqrt{x^{2} + 1} \cdot 2 \sqrt{x + 13}$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.5.3.1

Multiply $\frac{x}{\sqrt{x^{2} + 1}}$ by $\frac{2 \sqrt{x + 13}}{2 \sqrt{x + 13}}$ .

$lim_{x \to 4} \frac{\frac{x (2 \sqrt{x + 13})}{\sqrt{x^{2} + 1} (2 \sqrt{x + 13})} - \frac{1}{2 \sqrt{x + 13}} \cdot \frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}}{1}$

Step 1.5.3.2

Combine using the product rule for radicals.

$lim_{x \to 4} \frac{\frac{x (2 \sqrt{x + 13})}{2 \sqrt{(x^{2} + 1) (x + 13)}} - \frac{1}{2 \sqrt{x + 13}} \cdot \frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}}{1}$

Step 1.5.3.3

Multiply $\frac{1}{2 \sqrt{x + 13}}$ by $\frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}$ .

$lim_{x \to 4} \frac{\frac{x (2 \sqrt{x + 13})}{2 \sqrt{(x^{2} + 1) (x + 13)}} - \frac{\sqrt{x^{2} + 1}}{2 \sqrt{x + 13} \sqrt{x^{2} + 1}}}{1}$

Step 1.5.3.4

Combine using the product rule for radicals.

$lim_{x \to 4} \frac{\frac{x (2 \sqrt{x + 13})}{2 \sqrt{(x^{2} + 1) (x + 13)}} - \frac{\sqrt{x^{2} + 1}}{2 \sqrt{(x^{2} + 1) (x + 13)}}}{1}$

Step 1.5.4

Combine the numerators over the common denominator.

$lim_{x \to 4} \frac{\frac{x (2 \sqrt{x + 13}) - \sqrt{x^{2} + 1}}{2 \sqrt{(x^{2} + 1) (x + 13)}}}{1}$

Step 1.6

Divide $\frac{x (2 \sqrt{x + 13}) - \sqrt{x^{2} + 1}}{2 \sqrt{(x^{2} + 1) (x + 13)}}$ by $1$ .

$lim_{x \to 4} \frac{x (2 \sqrt{x + 13}) - \sqrt{x^{2} + 1}}{2 \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} lim_{x \to 4} \frac{x (2 \sqrt{x + 13}) - \sqrt{x^{2} + 1}}{\sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{lim_{x \to 4} x \cdot 2 \sqrt{x + 13} - \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{lim_{x \to 4} x \cdot 2 \sqrt{x + 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.4

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

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \sqrt{x + 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.5

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot lim_{x \to 4} \sqrt{x + 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.6

Move the limit under the radical sign.

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + lim_{x \to 4} 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.8

Evaluate the limit of $13$ which is constant as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - lim_{x \to 4} \sqrt{x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.9

Move the limit under the radical sign.

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{lim_{x \to 4} x^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.10

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{lim_{x \to 4} x^{2} + lim_{x \to 4} 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.11

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + lim_{x \to 4} 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.12

Evaluate the limit of $1$ which is constant as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{lim_{x \to 4} \sqrt{(x^{2} + 1) (x + 13)}}$

Step 2.13

Move the limit under the radical sign.

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{lim_{x \to 4} (x^{2} + 1) (x + 13)}}$

Step 2.14

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{lim_{x \to 4} x^{2} + 1 \cdot lim_{x \to 4} x + 13}}$

Step 2.15

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{(lim_{x \to 4} x^{2} + lim_{x \to 4} 1) \cdot lim_{x \to 4} x + 13}}$

Step 2.16

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + lim_{x \to 4} 1) \cdot lim_{x \to 4} x + 13}}$

Step 2.17

Evaluate the limit of $1$ which is constant as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot lim_{x \to 4} x + 13}}$

Step 2.18

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot (lim_{x \to 4} x + lim_{x \to 4} 13)}}$

Step 2.19

Evaluate the limit of $13$ which is constant as $x$ approaches $4$ .

$\frac{1}{2} \cdot \frac{2 lim_{x \to 4} x \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot (lim_{x \to 4} x + 13)}}$

Step 3

Evaluate the limits by plugging in $4$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2} \cdot \frac{2 \cdot 4 \cdot \sqrt{lim_{x \to 4} x + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot (lim_{x \to 4} x + 13)}}$

Step 3.2

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2} \cdot \frac{2 \cdot 4 \cdot \sqrt{4 + 13} - \sqrt{{(lim_{x \to 4} x)}^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot (lim_{x \to 4} x + 13)}}$

Step 3.3

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2} \cdot \frac{2 \cdot 4 \cdot \sqrt{4 + 13} - \sqrt{4^{2} + 1}}{\sqrt{({(lim_{x \to 4} x)}^{2} + 1) \cdot (lim_{x \to 4} x + 13)}}$

Step 3.4

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2} \cdot \frac{2 \cdot 4 \cdot \sqrt{4 + 13} - \sqrt{4^{2} + 1}}{\sqrt{(4^{2} + 1) \cdot (lim_{x \to 4} x + 13)}}$

Step 3.5

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2} \cdot \frac{2 \cdot 4 \cdot \sqrt{4 + 13} - \sqrt{4^{2} + 1}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $2$ by $4$ .

$\frac{1}{2} \cdot \frac{8 \cdot \sqrt{4 + 13} - \sqrt{4^{2} + 1}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4.1.2

Add $4$ and $13$ .

$\frac{1}{2} \cdot \frac{8 \cdot \sqrt{17} - \sqrt{4^{2} + 1}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4.1.3

Raise $4$ to the power of $2$ .

$\frac{1}{2} \cdot \frac{8 \sqrt{17} - \sqrt{16 + 1}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4.1.4

Add $16$ and $1$ .

$\frac{1}{2} \cdot \frac{8 \sqrt{17} - \sqrt{17}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4.1.5

Subtract $\sqrt{17}$ from $8 \sqrt{17}$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{(4^{2} + 1) \cdot (4 + 13)}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Raise $4$ to the power of $2$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{(16 + 1) (4 + 13)}}$

Step 4.2.2

Add $16$ and $1$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{17 (4 + 13)}}$

Step 4.2.3

Add $4$ and $13$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{17 \cdot 17}}$

Step 4.2.4

Multiply $17$ by $17$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{289}}$

Step 4.2.5

Rewrite $289$ as $17^{2}$ .

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{\sqrt{17^{2}}}$

Step 4.2.6

Pull terms out from under the radical, assuming positive real numbers.

$\frac{1}{2} \cdot \frac{7 \sqrt{17}}{17}$

Step 4.3

Multiply $\frac{1}{2} \cdot \frac{7 \sqrt{17}}{17}$ .

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Step 4.3.1

Multiply $\frac{1}{2}$ by $\frac{7 \sqrt{17}}{17}$ .

$\frac{7 \sqrt{17}}{2 \cdot 17}$

Step 4.3.2

Multiply $2$ by $17$ .

$\frac{7 \sqrt{17}}{34}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{7 \sqrt{17}}{34}$

Decimal Form:

$0.84887468 \dots$