Calculus Examples

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

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

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

Evaluate the limit.

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Simplify each term.

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

Cancel the common factor of $x^{6}$ .

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Cancel the common factor.

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

Divide $4$ by $1$ .

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

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

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

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Move the limit under the radical sign.

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

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

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

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{6}}$ approaches $0$ .

$\frac{- \sqrt{0 + lim_{x \to - \infty} 4}}{lim_{x \to - \infty} \frac{2}{x^{3}} - 1}$

Evaluate the limit.

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Evaluate the limit of $4$ which is constant as $x$ approaches $- \infty$ .

$\frac{- \sqrt{0 + 4}}{lim_{x \to - \infty} \frac{2}{x^{3}} - 1}$

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

$\frac{- \sqrt{0 + 4}}{lim_{x \to - \infty} \frac{2}{x^{3}} - lim_{x \to - \infty} 1}$

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

$\frac{- \sqrt{0 + 4}}{2 lim_{x \to - \infty} \frac{1}{x^{3}} - lim_{x \to - \infty} 1}$

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{3}}$ approaches $0$ .

$\frac{- \sqrt{0 + 4}}{2 \cdot 0 - lim_{x \to - \infty} 1}$

Evaluate the limit.

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

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

Simplify the answer.

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Simplify the numerator.

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Add $0$ and $4$ .

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

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

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

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

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

Simplify the denominator.

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Multiply $2$ by $0$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $0$ .

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

Multiply $- 1$ by $2$ .

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

Divide $- 2$ by $- 1$ .

$2$