Calculus Examples

$lim_{x \to - 3} \frac{x^{2} - 9}{\sqrt{x + 7} - 2}$

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.

$\frac{lim_{x \to - 3} x^{2} - 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{lim_{x \to - 3} x^{2} - lim_{x \to - 3} 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.1.2

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

$\frac{{(lim_{x \to - 3} x)}^{2} - lim_{x \to - 3} 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.1.3

Evaluate the limit of $9$ which is constant as $x$ approaches $- 3$ .

$\frac{{(lim_{x \to - 3} x)}^{2} - 1 \cdot 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{{(- 3)}^{2} - 1 \cdot 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Raise $- 3$ to the power of $2$ .

$\frac{9 - 1 \cdot 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.3.1.2

Multiply $- 1$ by $9$ .

$\frac{9 - 9}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

Step 1.1.2.3.2

Subtract $9$ from $9$ .

$\frac{0}{lim_{x \to - 3} \sqrt{x + 7} - 2}$

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 $- 3$ .

$\frac{0}{lim_{x \to - 3} \sqrt{x + 7} - lim_{x \to - 3} 2}$

Step 1.1.3.1.2

Move the limit under the radical sign.

$\frac{0}{\sqrt{lim_{x \to - 3} x + 7} - lim_{x \to - 3} 2}$

Step 1.1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{0}{\sqrt{lim_{x \to - 3} x + lim_{x \to - 3} 7} - lim_{x \to - 3} 2}$

Step 1.1.3.1.4

Evaluate the limit of $7$ which is constant as $x$ approaches $- 3$ .

$\frac{0}{\sqrt{lim_{x \to - 3} x + 7} - lim_{x \to - 3} 2}$

Step 1.1.3.1.5

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

$\frac{0}{\sqrt{lim_{x \to - 3} x + 7} - 1 \cdot 2}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{0}{\sqrt{- 3 + 7} - 1 \cdot 2}$

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 $- 3$ and $7$ .

$\frac{0}{\sqrt{4} - 1 \cdot 2}$

Step 1.1.3.3.1.2

Rewrite $4$ as $2^{2}$ .

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

Step 1.1.3.3.1.3

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

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

Step 1.1.3.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.1.3.3.2

Subtract $2$ from $2$ .

$\frac{0}{0}$

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.

$lim_{x \to - 3} \frac{x^{2} - 9}{\sqrt{x + 7} - 2} = lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} - 9]}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} - 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9]$ .

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3.3

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

$lim_{x \to - 3} \frac{2 x + \frac{d}{d x} [- 9]}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3.4

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

$lim_{x \to - 3} \frac{2 x + 0}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3.5

Add $2 x$ and $0$ .

$lim_{x \to - 3} \frac{2 x}{\frac{d}{d x} [\sqrt{x + 7} - 2]}$

Step 1.3.6

By the Sum Rule, the derivative of $\sqrt{x + 7} - 2$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x + 7}] + \frac{d}{d x} [- 2]$ .

$lim_{x \to - 3} \frac{2 x}{\frac{d}{d x} [\sqrt{x + 7}] + \frac{d}{d x} [- 2]}$

Step 1.3.7

Evaluate $\frac{d}{d x} [\sqrt{x + 7}]$ .

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Step 1.3.7.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 7}$ as ${(x + 7)}^{\frac{1}{2}}$ .

$lim_{x \to - 3} \frac{2 x}{\frac{d}{d x} [{(x + 7)}^{\frac{1}{2}}] + \frac{d}{d x} [- 2]}$

Step 1.3.7.2

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 $g (x) = x + 7$ .

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Step 1.3.7.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$lim_{x \to - 3} \frac{2 x}{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x + 7] + \frac{d}{d x} [- 2]}$

Step 1.3.7.2.2

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

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x + 7] + \frac{d}{d x} [- 2]}$

Step 1.3.7.2.3

Replace all occurrences of $u$ with $x + 7$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 7] + \frac{d}{d x} [- 2]}$

Step 1.3.7.3

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [7]) + \frac{d}{d x} [- 2]}$

Step 1.3.7.4

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

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [7]) + \frac{d}{d x} [- 2]}$

Step 1.3.7.5

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} - 1} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.6

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

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.7

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

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.8

Combine the numerators over the common denominator.

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1 - 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.9

Simplify the numerator.

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Step 1.3.7.9.1

Multiply $- 1$ by $2$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{1 - 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.9.2

Subtract $2$ from $1$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{\frac{- 1}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.10

Move the negative in front of the fraction.

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{- \frac{1}{2}} (1 + 0) + \frac{d}{d x} [- 2]}$

Step 1.3.7.11

Add $1$ and $0$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2} {(x + 7)}^{- \frac{1}{2}} \cdot 1 + \frac{d}{d x} [- 2]}$

Step 1.3.7.12

Combine $\frac{1}{2}$ and ${(x + 7)}^{- \frac{1}{2}}$ .

$lim_{x \to - 3} \frac{2 x}{\frac{{(x + 7)}^{- \frac{1}{2}}}{2} \cdot 1 + \frac{d}{d x} [- 2]}$

Step 1.3.7.13

Multiply $\frac{{(x + 7)}^{- \frac{1}{2}}}{2}$ by $1$ .

$lim_{x \to - 3} \frac{2 x}{\frac{{(x + 7)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [- 2]}$

Step 1.3.7.14

Move ${(x + 7)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2 {(x + 7)}^{\frac{1}{2}}} + \frac{d}{d x} [- 2]}$

Step 1.3.8

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

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2 {(x + 7)}^{\frac{1}{2}}} + 0}$

Step 1.3.9

Add $\frac{1}{2 {(x + 7)}^{\frac{1}{2}}}$ and $0$ .

$lim_{x \to - 3} \frac{2 x}{\frac{1}{2 {(x + 7)}^{\frac{1}{2}}}}$

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 3} 2 x (2 {(x + 7)}^{\frac{1}{2}})$

Step 1.5

Rewrite ${(x + 7)}^{\frac{1}{2}}$ as $\sqrt{x + 7}$ .

$lim_{x \to - 3} 2 x (2 \sqrt{x + 7})$

Step 1.6

Multiply $2$ by $2$ .

$lim_{x \to - 3} 4 x \sqrt{x + 7}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$4 lim_{x \to - 3} x \sqrt{x + 7}$

Step 2.2

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- 3$ .

$4 lim_{x \to - 3} x \cdot lim_{x \to - 3} \sqrt{x + 7}$

Step 2.3

Move the limit under the radical sign.

$4 lim_{x \to - 3} x \cdot \sqrt{lim_{x \to - 3} x + 7}$

Step 2.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$4 lim_{x \to - 3} x \cdot \sqrt{lim_{x \to - 3} x + lim_{x \to - 3} 7}$

Step 2.5

Evaluate the limit of $7$ which is constant as $x$ approaches $- 3$ .

$4 lim_{x \to - 3} x \cdot \sqrt{lim_{x \to - 3} x + 7}$

Step 3

Evaluate the limits by plugging in $- 3$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$4 \cdot - 3 \cdot \sqrt{lim_{x \to - 3} x + 7}$

Step 3.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$4 \cdot - 3 \cdot \sqrt{- 3 + 7}$

Step 4

Simplify the answer.

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Step 4.1

Multiply $4$ by $- 3$ .

$- 12 \cdot \sqrt{- 3 + 7}$

Step 4.2

Add $- 3$ and $7$ .

$- 12 \cdot \sqrt{4}$

Step 4.3

Rewrite $4$ as $2^{2}$ .

$- 12 \cdot \sqrt{2^{2}}$

Step 4.4

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

$- 12 \cdot 2$

Step 4.5

Multiply $- 12$ by $2$ .

$- 24$