# How do you find the limit of  [ x + sqrt (x^2 + 2x) ] as x approaches -oo?

Aug 15, 2016

$= - 1$

#### Explanation:

${\lim}_{x \to - \infty} \left[x + \sqrt{{x}^{2} + 2 x}\right]$

prepare for binomial expansion
${\lim}_{x \to - \infty} \left[x + | x | \sqrt{1 + \frac{2}{x}}\right]$

expand out
${\lim}_{x \to - \infty} \left[x + | x | \left(1 + \frac{1}{2} \frac{2}{x} + O \left(\frac{1}{x} ^ 2\right)\right)\right]$

${\lim}_{x \to - \infty} \left[x + | x | + | x \frac{|}{x} + | x | O \left(\frac{1}{x} ^ 2\right)\right]$

for $x < 0$

${\lim}_{x \to - \infty} \left[x + \left(- x\right) + \frac{- x}{x} + \left(- x\right) O \left(\frac{1}{x} ^ 2\right)\right]$

${\lim}_{x \to - \infty} \left[- 1 - x O \left(\frac{1}{x} ^ 2\right)\right]$

${\lim}_{x \to - \infty} \left[- 1 + O \left(\frac{1}{x}\right)\right]$

$= - 1$

Sep 20, 2016

To find this limit without binomial expansion see below.

#### Explanation:

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

Rewrite as a fraction and rationalize the numerator.

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

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

 = (-2x)/(x-sqrt(x^2(1+2/x)) $\text{ }$ for $x \ne 0$

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

Recall that $\sqrt{{x}^{2}} = \left\mid x \right\mid$ so for $x < 0$ we have $\sqrt{{x}^{2}} = - x$.

Our expression is equivalent to

$= \frac{- 2 x}{x + x \sqrt{1 + \frac{2}{x}}}$ $\text{ }$ for $x < 0$

$= \frac{- 2 x}{x \left(1 + \sqrt{1 + \frac{2}{x}}\right)}$ $\text{ }$ for $x < 0$

$= \frac{- 2}{1 + \sqrt{1 + \frac{2}{x}}}$ $\text{ }$ for $x < 0$

Evaluating the limit as $x \rightarrow - \infty$ yields

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