# Find the Limit

• Sep 18th 2009, 06:39 PM
yoman360
Find the Limit
Find the limit algebraically (without graphing)
$\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+4}}$

lim as x approaches negative infinity

• Sep 18th 2009, 07:02 PM
Krizalid
$\frac{x}{\sqrt{x^{2}+4}}=\frac{x}{-x\sqrt{1+\dfrac{4}{x^{2}}}}=-\frac{1}{\sqrt{1+\dfrac{4}{x^{2}}}},$

$|x|=-x$ since $x<0.$
• Sep 18th 2009, 07:06 PM
yoman360
Quote:

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

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

Uh thanks but im still confused (Speechless)
• Sep 18th 2009, 07:15 PM
yoman360
Quote:

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

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

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

how is absolute value of x = negative x?
• Sep 18th 2009, 07:19 PM
redsoxfan325
Quote:

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

how is absolute value of x = negative x?

I think what he meant to write is $\frac{x}{|x|}=-1$ since $x<0$.