Precalculus Examples

$lim_{x \to 8} \frac{\sqrt{9 + 2 x} - 5}{\sqrt[3]{x} - 2}$

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.

$\frac{lim_{x \to 8} \sqrt{9 + 2 x} - 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

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 $8$ .

$\frac{lim_{x \to 8} \sqrt{9 + 2 x} - lim_{x \to 8} 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.1.2

Move the limit under the radical sign.

$\frac{\sqrt{lim_{x \to 8} 9 + 2 x} - lim_{x \to 8} 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $8$ .

$\frac{\sqrt{lim_{x \to 8} 9 + lim_{x \to 8} 2 x} - lim_{x \to 8} 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.1.4

Evaluate the limit of $9$ which is constant as $x$ approaches $8$ .

$\frac{\sqrt{9 + lim_{x \to 8} 2 x} - lim_{x \to 8} 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.1.5

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

$\frac{\sqrt{9 + 2 lim_{x \to 8} x} - lim_{x \to 8} 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.1.6

Evaluate the limit of $5$ which is constant as $x$ approaches $8$ .

$\frac{\sqrt{9 + 2 lim_{x \to 8} x} - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $8$ for $x$ .

$\frac{\sqrt{9 + 2 \cdot 8} - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

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

Multiply $2$ by $8$ .

$\frac{\sqrt{9 + 16} - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.3.1.2

Add $9$ and $16$ .

$\frac{\sqrt{25} - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.3.1.3

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{5^{2}} - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.3.1.4

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

$\frac{5 - 1 \cdot 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.3.1.5

Multiply $- 1$ by $5$ .

$\frac{5 - 5}{lim_{x \to 8} \sqrt[3]{x} - 2}$

Step 1.1.2.3.2

Subtract $5$ from $5$ .

$\frac{0}{lim_{x \to 8} \sqrt[3]{x} - 2}$

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 $8$ .

$\frac{0}{lim_{x \to 8} \sqrt[3]{x} - lim_{x \to 8} 2}$

Step 1.1.3.1.2

Move the limit under the radical sign.

$\frac{0}{\sqrt[3]{lim_{x \to 8} x} - lim_{x \to 8} 2}$

Step 1.1.3.1.3

Evaluate the limit of $2$ which is constant as $x$ approaches $8$ .

$\frac{0}{\sqrt[3]{lim_{x \to 8} x} - 1 \cdot 2}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $8$ for $x$ .

$\frac{0}{\sqrt[3]{8} - 1 \cdot 2}$

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

Rewrite $8$ as $2^{3}$ .

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

Step 1.1.3.3.1.2

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

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

Step 1.1.3.3.1.3

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.

$lim_{x \to 8} \frac{\sqrt{9 + 2 x} - 5}{\sqrt[3]{x} - 2} = lim_{x \to 8} \frac{\frac{d}{d x} [\sqrt{9 + 2 x} - 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 8} \frac{\frac{d}{d x} [\sqrt{9 + 2 x} - 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.2

By the Sum Rule, the derivative of $\sqrt{9 + 2 x} - 5$ with respect to $x$ is $\frac{d}{d x} [\sqrt{9 + 2 x}] + \frac{d}{d x} [- 5]$ .

$lim_{x \to 8} \frac{\frac{d}{d x} [\sqrt{9 + 2 x}] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [\sqrt{9 + 2 x}]$ .

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Step 1.3.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{9 + 2 x}$ as ${(9 + 2 x)}^{\frac{1}{2}}$ .

$lim_{x \to 8} \frac{\frac{d}{d x} [{(9 + 2 x)}^{\frac{1}{2}}] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

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 $g (x) = 9 + 2 x$ .

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Step 1.3.3.2.1

To apply the Chain Rule, set $u$ as $9 + 2 x$ .

$lim_{x \to 8} \frac{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [9 + 2 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.2.2

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

$lim_{x \to 8} \frac{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [9 + 2 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.2.3

Replace all occurrences of $u$ with $9 + 2 x$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1} \frac{d}{d x} [9 + 2 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.3

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

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1} (\frac{d}{d x} [9] + \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.4

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

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1} (0 + \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.5

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1} (0 + 2 \frac{d}{d x} [x]) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.6

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

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.7

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

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.8

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

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.9

Combine the numerators over the common denominator.

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1 - 1 \cdot 2}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.10

Simplify the numerator.

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Step 1.3.3.10.1

Multiply $- 1$ by $2$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{1 - 2}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.10.2

Subtract $2$ from $1$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{\frac{- 1}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.11

Move the negative in front of the fraction.

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{- \frac{1}{2}} (0 + 2 \cdot 1) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.12

Multiply $2$ by $1$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{- \frac{1}{2}} (0 + 2) + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.13

Add $0$ and $2$ .

$lim_{x \to 8} \frac{\frac{1}{2} {(9 + 2 x)}^{- \frac{1}{2}} \cdot 2 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.14

Combine $\frac{1}{2}$ and ${(9 + 2 x)}^{- \frac{1}{2}}$ .

$lim_{x \to 8} \frac{\frac{{(9 + 2 x)}^{- \frac{1}{2}}}{2} \cdot 2 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.15

Combine $\frac{{(9 + 2 x)}^{- \frac{1}{2}}}{2}$ and $2$ .

$lim_{x \to 8} \frac{\frac{{(9 + 2 x)}^{- \frac{1}{2}} \cdot 2}{2} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.16

Move $2$ to the left of ${(9 + 2 x)}^{- \frac{1}{2}}$ .

$lim_{x \to 8} \frac{\frac{2 \cdot {(9 + 2 x)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.17

Move ${(9 + 2 x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 8} \frac{\frac{2}{2 {(9 + 2 x)}^{\frac{1}{2}}} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.18

Cancel the common factor.

$lim_{x \to 8} \frac{\frac{2}{2 {(9 + 2 x)}^{\frac{1}{2}}} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.3.19

Rewrite the expression.

$lim_{x \to 8} \frac{\frac{1}{{(9 + 2 x)}^{\frac{1}{2}}} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$lim_{x \to 8} \frac{\frac{1}{{(9 + 2 x)}^{\frac{1}{2}}} + 0}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.5

Simplify.

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Step 1.3.5.1

Add $\frac{1}{{(9 + 2 x)}^{\frac{1}{2}}}$ and $0$ .

$lim_{x \to 8} \frac{\frac{1}{{(9 + 2 x)}^{\frac{1}{2}}}}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.5.2

Reorder terms.

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [\sqrt[3]{x} - 2]}$

Step 1.3.6

By the Sum Rule, the derivative of $\sqrt[3]{x} - 2$ with respect to $x$ is $\frac{d}{d x} [\sqrt[3]{x}] + \frac{d}{d x} [- 2]$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [\sqrt[3]{x}] + \frac{d}{d x} [- 2]}$

Step 1.3.7

Evaluate $\frac{d}{d x} [\sqrt[3]{x}]$ .

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Step 1.3.7.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{d}{d x} [x^{\frac{1}{3}}] + \frac{d}{d x} [- 2]}$

Step 1.3.7.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = \frac{1}{3}$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{1}{3} - 1} + \frac{d}{d x} [- 2]}$

Step 1.3.7.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{1}{3} - 1 \cdot \frac{3}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.7.4

Combine $- 1$ and $\frac{3}{3}$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.7.5

Combine the numerators over the common denominator.

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{1 - 1 \cdot 3}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.7.6

Simplify the numerator.

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Step 1.3.7.6.1

Multiply $- 1$ by $3$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{1 - 3}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.7.6.2

Subtract $3$ from $1$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{\frac{- 2}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.7.7

Move the negative in front of the fraction.

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{- \frac{2}{3}} + \frac{d}{d x} [- 2]}$

Step 1.3.8

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

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} x^{- \frac{2}{3}} + 0}$

Step 1.3.9

Simplify.

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Step 1.3.9.1

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

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3} \cdot \frac{1}{x^{\frac{2}{3}}} + 0}$

Step 1.3.9.2

Combine terms.

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Step 1.3.9.2.1

Multiply $\frac{1}{3}$ by $\frac{1}{x^{\frac{2}{3}}}$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3 x^{\frac{2}{3}}} + 0}$

Step 1.3.9.2.2

Add $\frac{1}{3 x^{\frac{2}{3}}}$ and $0$ .

$lim_{x \to 8} \frac{\frac{1}{{(2 x + 9)}^{\frac{1}{2}}}}{\frac{1}{3 x^{\frac{2}{3}}}}$

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 8} \frac{1}{{(2 x + 9)}^{\frac{1}{2}}} (3 x^{\frac{2}{3}})$

Step 1.5

Rewrite ${(2 x + 9)}^{\frac{1}{2}}$ as $\sqrt{2 x + 9}$ .

$lim_{x \to 8} \frac{1}{\sqrt{2 x + 9}} (3 x^{\frac{2}{3}})$

Step 1.6

Combine factors.

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Step 1.6.1

Combine $3$ and $\frac{1}{\sqrt{2 x + 9}}$ .

$lim_{x \to 8} \frac{3}{\sqrt{2 x + 9}} x^{\frac{2}{3}}$

Step 1.6.2

Combine $\frac{3}{\sqrt{2 x + 9}}$ and $x^{\frac{2}{3}}$ .

$lim_{x \to 8} \frac{3 x^{\frac{2}{3}}}{\sqrt{2 x + 9}}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$3 lim_{x \to 8} \frac{x^{\frac{2}{3}}}{\sqrt{2 x + 9}}$

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $8$ .

$3 \frac{lim_{x \to 8} x^{\frac{2}{3}}}{lim_{x \to 8} \sqrt{2 x + 9}}$

Step 2.3

Move the exponent $\frac{2}{3}$ from $x^{\frac{2}{3}}$ outside the limit using the Limits Power Rule.

$3 \frac{{(lim_{x \to 8} x)}^{\frac{2}{3}}}{lim_{x \to 8} \sqrt{2 x + 9}}$

Step 2.4

Move the limit under the radical sign.

$3 \frac{{(lim_{x \to 8} x)}^{\frac{2}{3}}}{\sqrt{lim_{x \to 8} 2 x + 9}}$

Step 2.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $8$ .

$3 \frac{{(lim_{x \to 8} x)}^{\frac{2}{3}}}{\sqrt{lim_{x \to 8} 2 x + lim_{x \to 8} 9}}$

Step 2.6

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

$3 \frac{{(lim_{x \to 8} x)}^{\frac{2}{3}}}{\sqrt{2 lim_{x \to 8} x + lim_{x \to 8} 9}}$

Step 2.7

Evaluate the limit of $9$ which is constant as $x$ approaches $8$ .

$3 \frac{{(lim_{x \to 8} x)}^{\frac{2}{3}}}{\sqrt{2 lim_{x \to 8} x + 9}}$

Step 3

Evaluate the limits by plugging in $8$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $8$ for $x$ .

$3 \frac{8^{\frac{2}{3}}}{\sqrt{2 lim_{x \to 8} x + 9}}$

Step 3.2

Evaluate the limit of $x$ by plugging in $8$ for $x$ .

$3 \frac{8^{\frac{2}{3}}}{\sqrt{2 \cdot 8 + 9}}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Rewrite $8$ as $2^{3}$ .

$3 \frac{{(2^{3})}^{\frac{2}{3}}}{\sqrt{2 \cdot 8 + 9}}$

Step 4.1.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3 \frac{2^{3 (\frac{2}{3})}}{\sqrt{2 \cdot 8 + 9}}$

Step 4.1.3

Cancel the common factor of $3$ .

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Step 4.1.3.1

Cancel the common factor.

$3 \frac{2^{3 (\frac{2}{3})}}{\sqrt{2 \cdot 8 + 9}}$

Step 4.1.3.2

Rewrite the expression.

$3 \frac{2^{2}}{\sqrt{2 \cdot 8 + 9}}$

Step 4.1.4

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

$3 \frac{4}{\sqrt{2 \cdot 8 + 9}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $8$ .

$3 \frac{4}{\sqrt{16 + 9}}$

Step 4.2.2

Add $16$ and $9$ .

$3 \frac{4}{\sqrt{25}}$

Step 4.2.3

Rewrite $25$ as $5^{2}$ .

$3 \frac{4}{\sqrt{5^{2}}}$

Step 4.2.4

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

$3 (\frac{4}{5})$

Step 4.3

Multiply $3 (\frac{4}{5})$ .

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Step 4.3.1

Combine $3$ and $\frac{4}{5}$ .

$\frac{3 \cdot 4}{5}$

Step 4.3.2

Multiply $3$ by $4$ .

$\frac{12}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{12}{5}$

Decimal Form:

$2.4$