1. ## Find the Limit

Find the limit algebraically (without graphing)
$\displaystyle \lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}}$

lim as x approaches negative infinity

2. $\displaystyle \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},$

$\displaystyle |x|=-x$ since $\displaystyle x<0.$

3. Originally Posted by Krizalid
$\displaystyle \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},$

$\displaystyle |x|=-x$ since $\displaystyle x<0.$
Uh thanks but im still confused

4. Originally Posted by Krizalid
$\displaystyle \frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},$

$\displaystyle |x|=-x$ since $\displaystyle x<0.$
Actually when I looked it over and analyzed it more it makes sense except
$\displaystyle |x|=-x$ since $\displaystyle x<0.$

how is absolute value of x = negative x?

5. Originally Posted by yoman360
Actually when I looked it over and analyzed it more it makes sense except
$\displaystyle |x|=-x$ since $\displaystyle x<0.$

how is absolute value of x = negative x?
I think what he meant to write is $\displaystyle \frac{x}{|x|}=-1$ since $\displaystyle x<0$.