Find the Limit

**yoman360** · September 18th 2009, 06:39 PM

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

lim as x approaches negative infinity

I'm confused please help!

**Krizalid** · September 18th 2009, 07:02 PM

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

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

**yoman360** · September 18th 2009, 07:06 PM

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

**yoman360** · September 18th 2009, 07:15 PM

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?

**redsoxfan325** · September 18th 2009, 07:19 PM

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 $x<0$ .

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