Calculus Examples

$lim_{x \to \infty} \frac{x + 2}{\sqrt{16 x^{2} + 1}}$

Step 1

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x = \sqrt{x^{2}}$ .

$lim_{x \to \infty} \frac{\frac{x}{x} + \frac{2}{x}}{\sqrt{\frac{16 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$

Step 2

Evaluate the limit.

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Step 2.1

Cancel the common factor of $x$ .

$lim_{x \to \infty} \frac{1 + \frac{2}{x}}{\sqrt{\frac{16 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$

Step 2.2

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

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Step 2.2.1

Cancel the common factor.

$lim_{x \to \infty} \frac{1 + \frac{2}{x}}{\sqrt{\frac{16 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$

Step 2.2.2

Divide $16$ by $1$ .

$lim_{x \to \infty} \frac{1 + \frac{2}{x}}{\sqrt{16 + \frac{1}{x^{2}}}}$

Step 2.3

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

$\frac{lim_{x \to \infty} 1 + \frac{2}{x}}{lim_{x \to \infty} \sqrt{16 + \frac{1}{x^{2}}}}$

Step 2.4

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

$\frac{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{2}{x}}{lim_{x \to \infty} \sqrt{16 + \frac{1}{x^{2}}}}$

Step 2.5

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

$\frac{1 + lim_{x \to \infty} \frac{2}{x}}{lim_{x \to \infty} \sqrt{16 + \frac{1}{x^{2}}}}$

Step 2.6

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

$\frac{1 + 2 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} \sqrt{16 + \frac{1}{x^{2}}}}$

Step 3

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

$\frac{1 + 2 \cdot 0}{lim_{x \to \infty} \sqrt{16 + \frac{1}{x^{2}}}}$

Step 4

Evaluate the limit.

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Step 4.1

Move the limit under the radical sign.

$\frac{1 + 2 \cdot 0}{\sqrt{lim_{x \to \infty} 16 + \frac{1}{x^{2}}}}$

Step 4.2

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

$\frac{1 + 2 \cdot 0}{\sqrt{lim_{x \to \infty} 16 + lim_{x \to \infty} \frac{1}{x^{2}}}}$

Step 4.3

Evaluate the limit of $16$ which is constant as $x$ approaches $\infty$ .

$\frac{1 + 2 \cdot 0}{\sqrt{16 + lim_{x \to \infty} \frac{1}{x^{2}}}}$

Step 5

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

$\frac{1 + 2 \cdot 0}{\sqrt{16 + 0}}$

Step 6

Simplify the answer.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Multiply $2$ by $0$ .

$\frac{1 + 0}{\sqrt{16 + 0}}$

Step 6.1.2

Add $1$ and $0$ .

$\frac{1}{\sqrt{16 + 0}}$

Step 6.2

Simplify the denominator.

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Step 6.2.1

Add $16$ and $0$ .

$\frac{1}{\sqrt{16}}$

Step 6.2.2

Rewrite $16$ as $4^{2}$ .

$\frac{1}{\sqrt{4^{2}}}$

Step 6.2.3

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

$\frac{1}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{4}$

Decimal Form:

$0.25$