# Question #d42b9

Mar 5, 2016

${\lim}_{x \rightarrow - \infty} \left(\sqrt{{x}^{2} + 3 x} - x\right) = \infty$ and ${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 3 x} - x\right) = \frac{3}{2}$

#### Explanation:

In evaluating limits at infinity, we are not interested in what happens when $x = 0$, so we can write

$\sqrt{{x}^{2} + 3 x} = \sqrt{{x}^{2} \left(1 + \frac{3}{x}\right)} = \sqrt{{x}^{2}} \sqrt{1 + \frac{3}{x}}$

Recall that $\sqrt{{x}^{2}} = \left\mid x \right\mid$, so we may proceed:

${\lim}_{x \rightarrow - \infty} \left(\sqrt{{x}^{2} + 3 x} - x\right) = {\lim}_{x \rightarrow - \infty} \left(\sqrt{{x}^{2}} \sqrt{1 + \frac{3}{x}} - x\right)$

$= {\lim}_{x \rightarrow - \infty} \left(- x \sqrt{1 + \frac{3}{x}} - x\right)$

$= {\lim}_{x \rightarrow - \infty} \left(- x \left(\sqrt{1 + \frac{3}{x}} + 1\right)\right)$

$= - \left(- \infty\right) \left(2\right) = \infty$

(This result can be seen more quickly, if less formally, by observing that ar $x \rightarrow - \infty$, both $\sqrt{{x}^{2} + 3 x}$ and $- x$ increase without bound, so their sum increases without bound. That is, the initial form of the limit is $\infty - \left(- \infty\right)$ which gives a limit of $\infty$.)

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 3 x} - x\right)$ has initial form $\infty - \infty$ which is indeterminate.

$\left(\sqrt{{x}^{2} + 3 x} - x\right) = \frac{\sqrt{{x}^{2} + 3 x} - x}{1} \cdot \frac{\sqrt{{x}^{2} + 3 x} + x}{\sqrt{{x}^{2} + 3 x} + x}$

$= \frac{\left({x}^{2} + 3 x\right) - {x}^{2}}{\sqrt{{x}^{2} + 3 x} + x}$

$= \frac{3 x}{\sqrt{{x}^{2}} \sqrt{1 + \frac{3}{x}} + x}$

$= \frac{3 x}{x \sqrt{1 + \frac{3}{x}} + x}$ $\text{ }$ (for $x > 0$, we have $\sqrt{{x}^{2}} = x$)

$= \frac{3}{\sqrt{1 + \frac{3}{x}} + 1}$

${\lim}_{x \rightarrow \infty} \frac{3}{\sqrt{1 + \frac{3}{x}} + 1} = \frac{3}{2}$