# What is the limit of (sqrt(x^2+4x+1)-x)  as x goes to infinity?

Oct 10, 2015

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 4 x + 1} - x\right) = 2$

#### Explanation:

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 4 x + 1} - x\right)$ has form $\infty - \infty$ which is indeterminate.

$\left(\sqrt{{x}^{2} + 4 x + 1} - x\right) = \frac{\left(\sqrt{{x}^{2} + 4 x + 1} - x\right)}{1} \frac{\left(\sqrt{{x}^{2} + 4 x + 1} + x\right)}{\left(\sqrt{{x}^{2} + 4 x + 1} + x\right)}$

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

$= \frac{4 x + 1}{\sqrt{{x}^{2} \left(1 + \frac{4}{x} + \frac{1}{x} ^ 2\right)} + x}$ for $x \ne 0$

$= \frac{x \left(4 + \frac{1}{x}\right)}{x \left(\sqrt{1 + \frac{4}{x} + \frac{1}{x} ^ 2} + 1\right)}$ for $x > 0$

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

Taking the limit as $x \rightarrow \infty$, we get $\frac{4}{\sqrt{1} + 1} = \frac{4}{2} = 2$