Calculus Examples

$lim_{x \to - \infty} \frac{\sqrt[3]{x^{6} + 8}}{4 x^{2} + \sqrt{3 x^{4} + 1}}$

Step 1

Simplify.

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Step 1.1

Rewrite $x^{6}$ as ${(x^{2})}^{3}$ .

$lim_{x \to - \infty} \frac{\sqrt[3]{{(x^{2})}^{3} + 8}}{4 x^{2} + \sqrt{3 x^{4} + 1}}$

Step 1.2

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

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

Step 1.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x^{2}$ and $b = 2$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Multiply the exponents in ${(x^{2})}^{2}$ .

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Step 1.4.1.1

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

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

Step 1.4.1.2

Multiply $2$ by $2$ .

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

Step 1.4.2

Multiply $2$ by $- 1$ .

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

Step 1.4.3

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

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

Step 2

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $\sqrt[3]{x^{6}} = x^{2} = - \sqrt{x^{4}}$ .

$lim_{x \to - \infty} \frac{\sqrt[3]{\frac{(x^{2} + 2) (x^{4} - 2 x^{2} + 4)}{x^{6}}}}{\frac{4 x^{2}}{x^{2}} - \sqrt{\frac{3 x^{4}}{x^{4}} + \frac{1}{x^{4}}}}$

Step 3

Reduce the expression by cancelling the common factors.

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Step 3.1

Cancel the common factor of $x^{2}$ .

$lim_{x \to - \infty} \frac{\sqrt[3]{\frac{(x^{2} + 2) (x^{4} - 2 x^{2} + 4)}{x^{6}}}}{4 - \sqrt{\frac{3 x^{4}}{x^{4}} + \frac{1}{x^{4}}}}$

Step 3.2

Cancel the common factor of $x^{4}$ .

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Step 3.2.1

Cancel the common factor.

$lim_{x \to - \infty} \frac{\sqrt[3]{\frac{(x^{2} + 2) (x^{4} - 2 x^{2} + 4)}{x^{6}}}}{4 - \sqrt{\frac{3 x^{4}}{x^{4}} + \frac{1}{x^{4}}}}$

Step 3.2.2

Divide $3$ by $1$ .

$lim_{x \to - \infty} \frac{\sqrt[3]{\frac{(x^{2} + 2) (x^{4} - 2 x^{2} + 4)}{x^{6}}}}{4 - \sqrt{3 + \frac{1}{x^{4}}}}$

Step 4

Since $- \infty$ is not in the domain of $\frac{\sqrt[3]{\frac{(x^{2} + 2) (x^{4} - 2 x^{2} + 4)}{x^{6}}}}{4 - \sqrt{3 + \frac{1}{x^{4}}}}$ , the limit does not exist.

$Does not exist$