# How do you determine the limit of sqrt( x^2 + x) / (-2x) as n approaches -oo?

Apr 12, 2016

The limit is $\frac{1}{2}$. For "how" please see the explanation below.

#### Explanation:

Here are the crucial observations:

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

(The above is true provided that $x \ne 0$. But when evaluating limits as $x$ increases or decreases without bound, we aren't concerned with $x = 0$.)

Also observe/recall that for real $x$, we have

$\sqrt{{x}^{2}} = \left\mid x \right\mid = \left\{\begin{matrix}- x & \text{if" & x<0 \\ x & "if} & 0 \le x\end{matrix}\right.$.

Prepared with these two observations, we proceed:

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

$= {\lim}_{x \rightarrow - \infty} \frac{\left(- x\right) \sqrt{1 + \frac{1}{x}}}{- 2 x}$ $\text{ }$ [we have $x < 0$]

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

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

Bonus
Here is the limit as $x$ increases without bound. (I'm not a fan of the expression "approaches $\infty$".)

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

$= {\lim}_{x \rightarrow \infty} \frac{\left(x\right) \sqrt{1 + \frac{1}{x}}}{- 2 x}$ $\text{ }$ [we have $x > 0$]

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

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