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

[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!

 

[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

 

[dextercioby]

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

Daniel.