# How do you find the limit of x / sqrt (4x^2 + 2x +1)  as x approaches infinity?

Nov 29, 2016

#### Explanation:

$\frac{x}{\sqrt{4 {x}^{2} + 2 x + 1}} = \frac{x}{\sqrt{{x}^{2}} \sqrt{4 + \frac{2}{x} + \frac{1}{x} ^ 2}}$ $\text{ }$ for $x \ne 0$

$\sqrt{{x}^{2}} = \left\mid x \right\mid = x$ for $x > 0$

${\lim}_{x \rightarrow \infty} \frac{x}{\sqrt{4 {x}^{2} + 2 x + 1}} = {\lim}_{x \rightarrow \infty} \frac{x}{x \sqrt{4 + \frac{2}{x} + \frac{1}{x} ^ 2}}$

$= {\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{4 + \frac{2}{x} + \frac{1}{x} ^ 2}}$

$= \frac{1}{\sqrt{4 + 0 + 0}} = \frac{1}{2}$

Bonus material

Since $\sqrt{{x}^{2}} = - x$ for $x < 0$, we have

${\lim}_{x \rightarrow - \infty} \frac{x}{\sqrt{4 {x}^{2} + 2 x + 1}} = {\lim}_{x \rightarrow - \infty} \frac{x}{- x \sqrt{4 + \frac{2}{x} + \frac{1}{x} ^ 2}}$

$= {\lim}_{x \rightarrow - \infty} \frac{1}{- \sqrt{4 + \frac{2}{x} + \frac{1}{x} ^ 2}}$

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

Here is the graph

graph{x / sqrt (4x^2 + 2x +1) [-10, 10, -5, 5]}