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

Sep 16, 2015

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

#### Explanation:

(-2x+1)/sqrt(x^2 +x) = (x(-2+1/x))/(sqrt(x^2)sqrt(1+1/x)

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

When we find the limit as $x \rightarrow \infty$, we are concerned with positive values of $x$, so we have:

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

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

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

For limit as $x \rightarrow - \infty$

we use the fact the for negative values of $x$

$\sqrt{{x}^{2}} = \left\mid x \right\mid = - x$, to get

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

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

Here is the graph, so we can see the two horizontal asymptotes.

You can zoom in and out and drag the graph using a mouse.

graph{(-2x+1)/sqrt(x^2 +x) [-25.9, 31.81, -14.4, 14.46]}