Calculus Examples

$lim_{x \to 5} \frac{x^{2} - 5 x}{\sqrt{x + 4} - 3}$

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.

$\frac{lim_{x \to 5} x^{2} - 5 x}{lim_{x \to 5} \sqrt{x + 4} - 3}$

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 $5$ .

$\frac{lim_{x \to 5} x^{2} - lim_{x \to 5} 5 x}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to 5} x)}^{2} - lim_{x \to 5} 5 x}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.3

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{{(lim_{x \to 5} x)}^{2} - 5 lim_{x \to 5} x}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.4

Evaluate the limits by plugging in $5$ for all occurrences of $x$ .

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Step 1.1.2.4.1

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{5^{2} - 5 lim_{x \to 5} x}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.4.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{5^{2} - 5 \cdot 5}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.5

Simplify the answer.

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Step 1.1.2.5.1

Simplify each term.

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Step 1.1.2.5.1.1

Raise $5$ to the power of $2$ .

$\frac{25 - 5 \cdot 5}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.5.1.2

Multiply $- 5$ by $5$ .

$\frac{25 - 25}{lim_{x \to 5} \sqrt{x + 4} - 3}$

Step 1.1.2.5.2

Subtract $25$ from $25$ .

$\frac{0}{lim_{x \to 5} \sqrt{x + 4} - 3}$

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 $5$ .

$\frac{0}{lim_{x \to 5} \sqrt{x + 4} - lim_{x \to 5} 3}$

Step 1.1.3.1.2

Move the limit under the radical sign.

$\frac{0}{\sqrt{lim_{x \to 5} x + 4} - lim_{x \to 5} 3}$

Step 1.1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{0}{\sqrt{lim_{x \to 5} x + lim_{x \to 5} 4} - lim_{x \to 5} 3}$

Step 1.1.3.1.4

Evaluate the limit of $4$ which is constant as $x$ approaches $5$ .

$\frac{0}{\sqrt{lim_{x \to 5} x + 4} - lim_{x \to 5} 3}$

Step 1.1.3.1.5

Evaluate the limit of $3$ which is constant as $x$ approaches $5$ .

$\frac{0}{\sqrt{lim_{x \to 5} x + 4} - 1 \cdot 3}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{0}{\sqrt{5 + 4} - 1 \cdot 3}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Add $5$ and $4$ .

$\frac{0}{\sqrt{9} - 1 \cdot 3}$

Step 1.1.3.3.1.2

Rewrite $9$ as $3^{2}$ .

$\frac{0}{\sqrt{3^{2}} - 1 \cdot 3}$

Step 1.1.3.3.1.3

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

$\frac{0}{3 - 1 \cdot 3}$

Step 1.1.3.3.1.4

Multiply $- 1$ by $3$ .

$\frac{0}{3 - 3}$

Step 1.1.3.3.2

Subtract $3$ from $3$ .

$\frac{0}{0}$

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.

$lim_{x \to 5} \frac{x^{2} - 5 x}{\sqrt{x + 4} - 3} = lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 5 x]}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 5 x]}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} - 5 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x]$ .

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.3

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

$lim_{x \to 5} \frac{2 x + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 1.3.4.1

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

$lim_{x \to 5} \frac{2 x - 5 \frac{d}{d x} [x]}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.4.2

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

$lim_{x \to 5} \frac{2 x - 5 \cdot 1}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.4.3

Multiply $- 5$ by $1$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{d}{d x} [\sqrt{x + 4} - 3]}$

Step 1.3.5

By the Sum Rule, the derivative of $\sqrt{x + 4} - 3$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x + 4}] + \frac{d}{d x} [- 3]$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{d}{d x} [\sqrt{x + 4}] + \frac{d}{d x} [- 3]}$

Step 1.3.6

Evaluate $\frac{d}{d x} [\sqrt{x + 4}]$ .

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Step 1.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{d}{d x} [{(x + 4)}^{\frac{1}{2}}] + \frac{d}{d x} [- 3]}$

Step 1.3.6.2

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 $g (x) = x + 4$ .

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Step 1.3.6.2.1

To apply the Chain Rule, set $u$ as $x + 4$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x + 4] + \frac{d}{d x} [- 3]}$

Step 1.3.6.2.2

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x + 4] + \frac{d}{d x} [- 3]}$

Step 1.3.6.2.3

Replace all occurrences of $u$ with $x + 4$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 4] + \frac{d}{d x} [- 3]}$

Step 1.3.6.3

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [4]) + \frac{d}{d x} [- 3]}$

Step 1.3.6.4

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [4]) + \frac{d}{d x} [- 3]}$

Step 1.3.6.5

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} - 1} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.6

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.7

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.8

Combine the numerators over the common denominator.

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1 - 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.9

Simplify the numerator.

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Step 1.3.6.9.1

Multiply $- 1$ by $2$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{1 - 2}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.9.2

Subtract $2$ from $1$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{\frac{- 1}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.10

Move the negative in front of the fraction.

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{- \frac{1}{2}} (1 + 0) + \frac{d}{d x} [- 3]}$

Step 1.3.6.11

Add $1$ and $0$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2} {(x + 4)}^{- \frac{1}{2}} \cdot 1 + \frac{d}{d x} [- 3]}$

Step 1.3.6.12

Combine $\frac{1}{2}$ and ${(x + 4)}^{- \frac{1}{2}}$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{{(x + 4)}^{- \frac{1}{2}}}{2} \cdot 1 + \frac{d}{d x} [- 3]}$

Step 1.3.6.13

Multiply $\frac{{(x + 4)}^{- \frac{1}{2}}}{2}$ by $1$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{{(x + 4)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [- 3]}$

Step 1.3.6.14

Move ${(x + 4)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2 {(x + 4)}^{\frac{1}{2}}} + \frac{d}{d x} [- 3]}$

Step 1.3.7

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

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2 {(x + 4)}^{\frac{1}{2}}} + 0}$

Step 1.3.8

Add $\frac{1}{2 {(x + 4)}^{\frac{1}{2}}}$ and $0$ .

$lim_{x \to 5} \frac{2 x - 5}{\frac{1}{2 {(x + 4)}^{\frac{1}{2}}}}$

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 5} (2 x - 5) (2 {(x + 4)}^{\frac{1}{2}})$

Step 1.5

Rewrite ${(x + 4)}^{\frac{1}{2}}$ as $\sqrt{x + 4}$ .

$lim_{x \to 5} (2 x - 5) (2 \sqrt{x + 4})$

Step 2

Evaluate the limit.

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Step 2.1

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

$2 lim_{x \to 5} (2 x - 5) \sqrt{x + 4}$

Step 2.2

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $5$ .

$2 lim_{x \to 5} 2 x - 5 \cdot lim_{x \to 5} \sqrt{x + 4}$

Step 2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$2 (lim_{x \to 5} 2 x - lim_{x \to 5} 5) \cdot lim_{x \to 5} \sqrt{x + 4}$

Step 2.4

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

$2 (2 lim_{x \to 5} x - lim_{x \to 5} 5) \cdot lim_{x \to 5} \sqrt{x + 4}$

Step 2.5

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

$2 (2 lim_{x \to 5} x - 1 \cdot 5) \cdot lim_{x \to 5} \sqrt{x + 4}$

Step 2.6

Move the limit under the radical sign.

$2 (2 lim_{x \to 5} x - 1 \cdot 5) \cdot \sqrt{lim_{x \to 5} x + 4}$

Step 2.7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$2 (2 lim_{x \to 5} x - 1 \cdot 5) \cdot \sqrt{lim_{x \to 5} x + lim_{x \to 5} 4}$

Step 2.8

Evaluate the limit of $4$ which is constant as $x$ approaches $5$ .

$2 (2 lim_{x \to 5} x - 1 \cdot 5) \cdot \sqrt{lim_{x \to 5} x + 4}$

Step 3

Evaluate the limits by plugging in $5$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$2 (2 \cdot 5 - 1 \cdot 5) \cdot \sqrt{lim_{x \to 5} x + 4}$

Step 3.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$2 (2 \cdot 5 - 1 \cdot 5) \cdot \sqrt{5 + 4}$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $2$ by $5$ .

$2 (10 - 1 \cdot 5) \cdot \sqrt{5 + 4}$

Step 4.1.2

Multiply $- 1$ by $5$ .

$2 (10 - 5) \cdot \sqrt{5 + 4}$

Step 4.2

Subtract $5$ from $10$ .

$2 \cdot 5 \cdot \sqrt{5 + 4}$

Step 4.3

Multiply $2$ by $5$ .

$10 \cdot \sqrt{5 + 4}$

Step 4.4

Add $5$ and $4$ .

$10 \cdot \sqrt{9}$

Step 4.5

Rewrite $9$ as $3^{2}$ .

$10 \cdot \sqrt{3^{2}}$

Step 4.6

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

$10 \cdot 3$

Step 4.7

Multiply $10$ by $3$ .

$30$