Calculus Examples

$lim_{x \to 9} \frac{x + \sqrt{x} - 12}{x - 9}$

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.

$\frac{lim_{x \to 9} x + \sqrt{x} - 12}{lim_{x \to 9} x - 9}$

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 $9$ .

$\frac{lim_{x \to 9} x + lim_{x \to 9} \sqrt{x} - lim_{x \to 9} 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.2

Move the limit under the radical sign.

$\frac{lim_{x \to 9} x + \sqrt{lim_{x \to 9} x} - lim_{x \to 9} 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.3

Evaluate the limit of $12$ which is constant as $x$ approaches $9$ .

$\frac{lim_{x \to 9} x + \sqrt{lim_{x \to 9} x} - 1 \cdot 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.4

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

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Step 1.1.2.4.1

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

$\frac{9 + \sqrt{lim_{x \to 9} x} - 1 \cdot 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.4.2

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

$\frac{9 + \sqrt{9} - 1 \cdot 12}{lim_{x \to 9} x - 9}$

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

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

$\frac{9 + \sqrt{3^{2}} - 1 \cdot 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.5.1.2

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

$\frac{9 + 3 - 1 \cdot 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.5.1.3

Multiply $- 1$ by $12$ .

$\frac{9 + 3 - 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.5.2

Add $9$ and $3$ .

$\frac{12 - 12}{lim_{x \to 9} x - 9}$

Step 1.1.2.5.3

Subtract $12$ from $12$ .

$\frac{0}{lim_{x \to 9} x - 9}$

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 $9$ .

$\frac{0}{lim_{x \to 9} x - lim_{x \to 9} 9}$

Step 1.1.3.1.2

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

$\frac{0}{lim_{x \to 9} x - 1 \cdot 9}$

Step 1.1.3.2

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

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

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $- 1$ by $9$ .

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

Step 1.1.3.3.2

Subtract $9$ from $9$ .

$\frac{0}{0}$

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.

$lim_{x \to 9} \frac{x + \sqrt{x} - 12}{x - 9} = lim_{x \to 9} \frac{\frac{d}{d x} [x + \sqrt{x} - 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 9} \frac{\frac{d}{d x} [x + \sqrt{x} - 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.2

By the Sum Rule, the derivative of $x + \sqrt{x} - 12$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- 12]$ .

$lim_{x \to 9} \frac{\frac{d}{d x} [x] + \frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.3

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

$lim_{x \to 9} \frac{1 + \frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4

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

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Step 1.3.4.1

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

$lim_{x \to 9} \frac{1 + \frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

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 $n = \frac{1}{2}$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{1}{2} - 1} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.3

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

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.4

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

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.5

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.6

Simplify the numerator.

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Step 1.3.4.6.1

Multiply $- 1$ by $2$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{1 - 2}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.6.2

Subtract $2$ from $1$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{\frac{- 1}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.4.7

Move the negative in front of the fraction.

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{- \frac{1}{2}} + \frac{d}{d x} [- 12]}{\frac{d}{d x} [x - 9]}$

Step 1.3.5

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2} x^{- \frac{1}{2}} + 0}{\frac{d}{d x} [x - 9]}$

Step 1.3.6

Simplify.

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Step 1.3.6.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2} \cdot \frac{1}{x^{\frac{1}{2}}} + 0}{\frac{d}{d x} [x - 9]}$

Step 1.3.6.2

Combine terms.

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Step 1.3.6.2.1

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}} + 0}{\frac{d}{d x} [x - 9]}$

Step 1.3.6.2.2

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}}}{\frac{d}{d x} [x - 9]}$

Step 1.3.7

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}}}{\frac{d}{d x} [x] + \frac{d}{d x} [- 9]}$

Step 1.3.8

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}}}{1 + \frac{d}{d x} [- 9]}$

Step 1.3.9

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}}}{1 + 0}$

Step 1.3.10

Add $1$ and $0$ .

$lim_{x \to 9} \frac{1 + \frac{1}{2 x^{\frac{1}{2}}}}{1}$

Step 1.4

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

$lim_{x \to 9} \frac{1 + \frac{1}{2 \sqrt{x}}}{1}$

Step 1.5

Combine terms.

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Step 1.5.1

Write $1$ as a fraction with a common denominator.

$lim_{x \to 9} \frac{\frac{2 \sqrt{x}}{2 \sqrt{x}} + \frac{1}{2 \sqrt{x}}}{1}$

Step 1.5.2

Combine the numerators over the common denominator.

$lim_{x \to 9} \frac{\frac{2 \sqrt{x} + 1}{2 \sqrt{x}}}{1}$

Step 1.6

Divide $\frac{2 \sqrt{x} + 1}{2 \sqrt{x}}$ by $1$ .

$lim_{x \to 9} \frac{2 \sqrt{x} + 1}{2 \sqrt{x}}$

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 $x$ .

$\frac{1}{2} lim_{x \to 9} \frac{2 \sqrt{x} + 1}{\sqrt{x}}$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Move the limit under the radical sign.

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

Step 2.6

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

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

Step 2.7

Move the limit under the radical sign.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$\frac{1}{2} \cdot \frac{2 \sqrt{9} + 1}{\sqrt{9}}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$\frac{1}{2} \cdot \frac{2 \sqrt{3^{2}} + 1}{\sqrt{9}}$

Step 4.1.2

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

$\frac{1}{2} \cdot \frac{2 \cdot 3 + 1}{\sqrt{9}}$

Step 4.1.3

Multiply $2$ by $3$ .

$\frac{1}{2} \cdot \frac{6 + 1}{\sqrt{9}}$

Step 4.1.4

Add $6$ and $1$ .

$\frac{1}{2} \cdot \frac{7}{\sqrt{9}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

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

$\frac{1}{2} \cdot \frac{7}{\sqrt{3^{2}}}$

Step 4.2.2

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

$\frac{1}{2} \cdot \frac{7}{3}$

Step 4.3

Multiply $\frac{1}{2} \cdot \frac{7}{3}$ .

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Step 4.3.1

Multiply $\frac{1}{2}$ by $\frac{7}{3}$ .

$\frac{7}{2 \cdot 3}$

Step 4.3.2

Multiply $2$ by $3$ .

$\frac{7}{6}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{7}{6}$

Decimal Form:

$1.1\overline{6}$