Simplify the Radical Expression limit as x approaches -2 of ( cube root of x-6+2)/(x^3+8)

$lim_{x \to - 2} \frac{\sqrt[3]{x - 6} + 2}{x^{3} + 8}$

Step 1

Apply L\'Hospital\'s rule.

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

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

$\frac{lim_{x \to - 2} \sqrt[3]{x - 6} + 2}{lim_{x \to - 2} x^{3} + 8}$

Evaluate the limit of the numerator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{lim_{x \to - 2} \sqrt[3]{x - 6} + lim_{x \to - 2} 2}{lim_{x \to - 2} x^{3} + 8}$

Move the limit under the radical sign.

$\frac{\sqrt[3]{lim_{x \to - 2} x - 6} + lim_{x \to - 2} 2}{lim_{x \to - 2} x^{3} + 8}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{\sqrt[3]{lim_{x \to - 2} x - lim_{x \to - 2} 6} + lim_{x \to - 2} 2}{lim_{x \to - 2} x^{3} + 8}$

Evaluate the limit of $6$ which is constant as $x$ approaches $- 2$ .

$\frac{\sqrt[3]{lim_{x \to - 2} x - 1 \cdot 6} + lim_{x \to - 2} 2}{lim_{x \to - 2} x^{3} + 8}$

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

$\frac{\sqrt[3]{lim_{x \to - 2} x - 1 \cdot 6} + 2}{lim_{x \to - 2} x^{3} + 8}$

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{\sqrt[3]{- 2 - 1 \cdot 6} + 2}{lim_{x \to - 2} x^{3} + 8}$

Simplify the answer.

Simplify each term.

Multiply $- 1$ by $6$ .

$\frac{\sqrt[3]{- 2 - 6} + 2}{lim_{x \to - 2} x^{3} + 8}$

Subtract $6$ from $- 2$ .

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

Rewrite $- 8$ as ${(- 2)}^{3}$ .

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

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

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

Add $- 2$ and $2$ .

$\frac{0}{lim_{x \to - 2} x^{3} + 8}$

Evaluate the limit of the denominator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{0}{lim_{x \to - 2} x^{3} + lim_{x \to - 2} 8}$

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

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

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

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

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{0}{{(- 2)}^{3} + 8}$

Simplify the answer.

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

$\frac{0}{- 8 + 8}$

Add $- 8$ and $8$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to - 2} \frac{\sqrt[3]{x - 6} + 2}{x^{3} + 8} = lim_{x \to - 2} \frac{\frac{d}{d x} [\sqrt[3]{x - 6} + 2]}{\frac{d}{d x} [x^{3} + 8]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

$lim_{x \to - 2} \frac{\frac{d}{d x} [\sqrt[3]{x - 6} + 2]}{\frac{d}{d x} [x^{3} + 8]}$

By the Sum Rule, the derivative of $\sqrt[3]{x - 6} + 2$ with respect to $x$ is $\frac{d}{d x} [\sqrt[3]{x - 6}] + \frac{d}{d x} [2]$ .

$lim_{x \to - 2} \frac{\frac{d}{d x} [\sqrt[3]{x - 6}] + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Evaluate $\frac{d}{d x} [\sqrt[3]{x - 6}]$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 6}$ as ${(x - 6)}^{\frac{1}{3}}$ .

$lim_{x \to - 2} \frac{\frac{d}{d x} [{(x - 6)}^{\frac{1}{3}}] + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{3}}$ and $g (x) = x - 6$ .

To apply the Chain Rule, set $u$ as $x - 6$ .

$lim_{x \to - 2} \frac{\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [x - 6] + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [x - 6] + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Replace all occurrences of $u$ with $x - 6$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} - 1} \frac{d}{d x} [x - 6] + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

By the Sum Rule, the derivative of $x - 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [- 6]) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} - 1} (1 + \frac{d}{d x} [- 6]) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} - 1} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} - 1 \cdot \frac{3}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Combine the numerators over the common denominator.

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1 - 1 \cdot 3}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Simplify the numerator.

Multiply $- 1$ by $3$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{1 - 3}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Subtract $3$ from $1$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{\frac{- 2}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Move the negative in front of the fraction.

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{- \frac{2}{3}} (1 + 0) + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Add $1$ and $0$ .

$lim_{x \to - 2} \frac{\frac{1}{3} {(x - 6)}^{- \frac{2}{3}} \cdot 1 + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Combine $\frac{1}{3}$ and ${(x - 6)}^{- \frac{2}{3}}$ .

$lim_{x \to - 2} \frac{\frac{{(x - 6)}^{- \frac{2}{3}}}{3} \cdot 1 + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Multiply $\frac{{(x - 6)}^{- \frac{2}{3}}}{3}$ by $1$ .

$lim_{x \to - 2} \frac{\frac{{(x - 6)}^{- \frac{2}{3}}}{3} + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

Move ${(x - 6)}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}} + \frac{d}{d x} [2]}{\frac{d}{d x} [x^{3} + 8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}} + 0}{\frac{d}{d x} [x^{3} + 8]}$

Add $\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}$ and $0$ .

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}}{\frac{d}{d x} [x^{3} + 8]}$

By the Sum Rule, the derivative of $x^{3} + 8$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [8]$ .

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}}{\frac{d}{d x} [x^{3}] + \frac{d}{d x} [8]}$

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

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}}{3 x^{2} + \frac{d}{d x} [8]}$

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}}{3 x^{2} + 0}$

Add $3 x^{2}$ and $0$ .

$lim_{x \to - 2} \frac{\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}}{3 x^{2}}$

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 2} \frac{1}{3 {(x - 6)}^{\frac{2}{3}}} \cdot \frac{1}{3 x^{2}}$

Combine factors.

Multiply $\frac{1}{3 {(x - 6)}^{\frac{2}{3}}}$ by $\frac{1}{3 x^{2}}$ .

$lim_{x \to - 2} \frac{1}{3 {(x - 6)}^{\frac{2}{3}} (3 x^{2})}$

Multiply $3$ by $3$ .

$lim_{x \to - 2} \frac{1}{9 {(x - 6)}^{\frac{2}{3}} x^{2}}$

Step 2

Move the term $\frac{1}{9}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{9} lim_{x \to - 2} \frac{1}{{(x - 6)}^{\frac{2}{3}} x^{2}}$

Step 3

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 2$ .

$\frac{1}{9} \cdot \frac{lim_{x \to - 2} 1}{lim_{x \to - 2} {(x - 6)}^{\frac{2}{3}} x^{2}}$

Step 4

Evaluate the limit of $1$ which is constant as $x$ approaches $- 2$ .

$\frac{1}{9} \cdot \frac{1}{lim_{x \to - 2} {(x - 6)}^{\frac{2}{3}} x^{2}}$

Step 5

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{1}{9} \cdot \frac{1}{lim_{x \to - 2} {(x - 6)}^{\frac{2}{3}} \cdot lim_{x \to - 2} x^{2}}$

Step 6

Move the exponent $\frac{2}{3}$ from ${(x - 6)}^{\frac{2}{3}}$ outside the limit using the Limits Power Rule.

$\frac{1}{9} \cdot \frac{1}{{(lim_{x \to - 2} x - 6)}^{\frac{2}{3}} \cdot lim_{x \to - 2} x^{2}}$

Step 7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 2$ .

$\frac{1}{9} \cdot \frac{1}{{(lim_{x \to - 2} x - lim_{x \to - 2} 6)}^{\frac{2}{3}} \cdot lim_{x \to - 2} x^{2}}$

Step 8

Evaluate the limit of $6$ which is constant as $x$ approaches $- 2$ .

$\frac{1}{9} \cdot \frac{1}{{(lim_{x \to - 2} x - 1 \cdot 6)}^{\frac{2}{3}} \cdot lim_{x \to - 2} x^{2}}$

Step 9

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

$\frac{1}{9} \cdot \frac{1}{{(lim_{x \to - 2} x - 1 \cdot 6)}^{\frac{2}{3}} \cdot {(lim_{x \to - 2} x)}^{2}}$

Step 10

Combine.

$\frac{1 \cdot 1}{9 ({(lim_{x \to - 2} x - 1 \cdot 6)}^{\frac{2}{3}} \cdot {(lim_{x \to - 2} x)}^{2})}$

Step 11

Multiply $1$ by $1$ .

$\frac{1}{9 {(lim_{x \to - 2} x - 1 \cdot 6)}^{\frac{2}{3}} \cdot {lim_{x \to - 2} x}^{2}}$

Step 12

Simplify the denominator.

Multiply $- 1$ by $6$ .

$\frac{1}{9 {(lim_{x \to - 2} x - 6)}^{\frac{2}{3}} {lim_{x \to - 2} x}^{2}}$

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{1}{9 {(- 2 - 6)}^{\frac{2}{3}} {lim_{x \to - 2} x}^{2}}$

Subtract $6$ from $- 2$ .

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

Evaluate the limit of $x$ by plugging in $- 2$ for $x$ .

$\frac{1}{9 {(- 8)}^{\frac{2}{3}} {(- 2)}^{2}}$

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

$\frac{1}{9 {(- 8)}^{\frac{2}{3}} \cdot 4}$

Multiply $4$ by $9$ .

$\frac{1}{36 {(- 8)}^{\frac{2}{3}}}$

Rewrite $- 8$ as ${(- 2)}^{3}$ .

$\frac{1}{36 {({(- 2)}^{3})}^{\frac{2}{3}}}$

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

$\frac{1}{36 {(- 2)}^{3 (\frac{2}{3})}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{1}{36 {(- 2)}^{3 (\frac{2}{3})}}$

Rewrite the expression.

$\frac{1}{36 {(- 2)}^{2}}$

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

$\frac{1}{36 \cdot 4}$

Step 13

Multiply $36$ by $4$ .

$\frac{1}{144}$

Simplify the Radical Expression limit as x approaches -2 of ( cube root of x-6+2)/(x^3+8)

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