Limit x->-infinity (x/(z^2+x^2)^0.5) = ?

1. Jun 22, 2005

Eus

Hi Ho!

Mmmm... I have a problem with this one:

$\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}}$

Using a computer graphics tool, I found that the result should be -1 by looking at the generated graph.

But if I do it by hands, I find 1 as follows:

$\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}} = \lim_{x\rightarrow-\infty} \frac{x \frac{1}{x}}{\sqrt{z^2+x^2} \frac{1}{x}}$
$= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{\frac{z^2+x^2}{x^2}}}$
$= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{1+(\frac{z}{x})^2}}$
$= \frac{1}{\sqrt{1+(\frac{z}{-\infty})^2}}$
$= \frac{1}{\sqrt{1+0}}$
$= \frac{1}{1}$
$= 1$

Would you please correct my mistake?

Thank you very much!

Last edited: Jun 22, 2005
2. Jun 22, 2005

steven187

hello there

check this out

$\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}}$
$=\lim_{n\rightarrow\infty} \frac{-n}{\sqrt{k^2+n^2}}$
$= \lim_{n\rightarrow\infty} \frac{-n\frac{1}{n}}{\sqrt{k^2+n^2} \frac{1}{n}}$
$= \lim_{n\rightarrow\infty} \frac{-1}{\sqrt{\frac{k^2+n^2}{n^2}}}$
$= \lim_{n\rightarrow\infty} \frac{-1}{\sqrt{1+(\frac{k}{n})^2}}$
$= \frac{-1}{\sqrt{1+0}}=-1$

always try to simplify it to something that looks simple

take care

steven

3. Jun 22, 2005

dextercioby

I hope you saw your mistake,you shouln't have sent "x" into its square when taking the limit to "-infinity".

Daniel.