Let $\displaystyle f(x) = |x|^{-1/2}$ for $\displaystyle x\neq0$. I need to show that $\displaystyle \lim_{x\to0+} f(x)= \lim_{x\to0-} f(x)=+\infty$...

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- Feb 9th 2010, 04:50 PMCrazyCat87Limits
Let $\displaystyle f(x) = |x|^{-1/2}$ for $\displaystyle x\neq0$. I need to show that $\displaystyle \lim_{x\to0+} f(x)= \lim_{x\to0-} f(x)=+\infty$...

- Feb 9th 2010, 09:31 PMmiguemate
Hello, let me try to help you

1) Try when x<0, then |x|=-x.

2) Try when x>0, then |x|=x.

Take the limit, and tell what you get.

Hugs - Feb 10th 2010, 12:04 AMxalk
- Feb 10th 2010, 05:35 AMHallsofIvy
- Feb 10th 2010, 06:57 AMCrazyCat87
- Feb 10th 2010, 07:52 AMHallsofIvy
Because making the substitution y= -x in $\displaystyle \lim_{x\to -\infty} |x|^{-1/2}$ gives $\displaystyle \lim_{y\to\infty} |-y|^{-1/2}= \lim_{y\to\infty} |y|^{-1/2}$ which is, of course, the same as $\displaystyle \lim_{x\to \infty}|x|^{-1/2}$

- Feb 10th 2010, 11:24 AMCrazyCat87
ok I'm done proving it, thanks everyone for all your help!!