Calculus Examples

$lim_{x \to \infty} \frac{x^{3} - x + 7}{\sqrt{9 x^{6} + 1}}$

Step 1

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

$lim_{x \to \infty} \frac{\frac{x^{3}}{x^{3}} - \frac{x}{x^{3}} + \frac{7}{x^{3}}}{\sqrt{\frac{9 x^{6}}{x^{6}} + \frac{1}{x^{6}}}}$

Step 2

Evaluate the limit.

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Step 2.1

Simplify each term.

$lim_{x \to \infty} \frac{1 - \frac{1}{x^{2}} + \frac{7}{x^{3}}}{\sqrt{\frac{9 x^{6}}{x^{6}} + \frac{1}{x^{6}}}}$

Step 2.2

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

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Step 2.2.1

Cancel the common factor.

$lim_{x \to \infty} \frac{1 - \frac{1}{x^{2}} + \frac{7}{x^{3}}}{\sqrt{\frac{9 x^{6}}{x^{6}} + \frac{1}{x^{6}}}}$

Step 2.2.2

Divide $9$ by $1$ .

$lim_{x \to \infty} \frac{1 - \frac{1}{x^{2}} + \frac{7}{x^{3}}}{\sqrt{9 + \frac{1}{x^{6}}}}$

Step 2.3

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

$\frac{lim_{x \to \infty} 1 - \frac{1}{x^{2}} + \frac{7}{x^{3}}}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 2.4

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

$\frac{lim_{x \to \infty} 1 - lim_{x \to \infty} \frac{1}{x^{2}} + lim_{x \to \infty} \frac{7}{x^{3}}}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 2.5

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

$\frac{1 - lim_{x \to \infty} \frac{1}{x^{2}} + lim_{x \to \infty} \frac{7}{x^{3}}}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

$\frac{1 - 0 + lim_{x \to \infty} \frac{7}{x^{3}}}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 4

Move the term $7$ outside of the limit because it is constant with respect to $x$ .

$\frac{1 - 0 + 7 lim_{x \to \infty} \frac{1}{x^{3}}}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 5

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

$\frac{1 - 0 + 7 \cdot 0}{lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{6}}}}$

Step 6

Evaluate the limit.

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Step 6.1

Move the limit under the radical sign.

$\frac{1 - 0 + 7 \cdot 0}{\sqrt{lim_{x \to \infty} 9 + \frac{1}{x^{6}}}}$

Step 6.2

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

$\frac{1 - 0 + 7 \cdot 0}{\sqrt{lim_{x \to \infty} 9 + lim_{x \to \infty} \frac{1}{x^{6}}}}$

Step 6.3

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

$\frac{1 - 0 + 7 \cdot 0}{\sqrt{9 + lim_{x \to \infty} \frac{1}{x^{6}}}}$

Step 7

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

$\frac{1 - 0 + 7 \cdot 0}{\sqrt{9 + 0}}$

Step 8

Simplify the answer.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Multiply $- 1$ by $0$ .

$\frac{1 + 0 + 7 \cdot 0}{\sqrt{9 + 0}}$

Step 8.1.2

Multiply $7$ by $0$ .

$\frac{1 + 0 + 0}{\sqrt{9 + 0}}$

Step 8.1.3

Add $1 + 0$ and $0$ .

$\frac{1 + 0}{\sqrt{9 + 0}}$

Step 8.1.4

Add $1$ and $0$ .

$\frac{1}{\sqrt{9 + 0}}$

Step 8.2

Simplify the denominator.

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Step 8.2.1

Add $9$ and $0$ .

$\frac{1}{\sqrt{9}}$

Step 8.2.2

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

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

Step 8.2.3

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

$\frac{1}{3}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{3}$

Decimal Form:

$0.\overline{3}$