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

  • Thread starter Eus
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Eus

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Hi Ho! :smile:

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

[itex]\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}}[/itex]

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:

[itex]\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}}[/itex]
[itex]= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{\frac{z^2+x^2}{x^2}}}[/itex]
[itex]= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{1+(\frac{z}{x})^2}}[/itex]
[itex]= \frac{1}{\sqrt{1+(\frac{z}{-\infty})^2}}[/itex]
[itex]= \frac{1}{\sqrt{1+0}}[/itex]
[itex]= \frac{1}{1}[/itex]
[itex]= 1[/itex]

Would you please correct my mistake?

Thank you very much! :biggrin:
 
Last edited:
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hello there

check this out

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

always try to simplify it to something that looks simple

take care

steven
 

dextercioby

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

Daniel.
 

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