# Limits

• Feb 9th 2010, 04:50 PM
CrazyCat87
Limits
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 PM
miguemate

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 AM
xalk
Quote:

Originally Posted by CrazyCat87
Let $\displaystyle f(x) = |x|^{-1/2}$ for $\displaystyle x\neq0$. I need to show that $\displaystyle \lim_{x\to0+} f(x)= \lim_{x\to0-} f(x)=+\infty$...

To prove that:

$\displaystyle lim_{x\to 0^+} f(x) =+\infty$ you must prove that:

for all ε>0 there exists a δ>0 such that :

for all ,x : if 0<x<δ , then f(x)>ε
• Feb 10th 2010, 05:35 AM
HallsofIvy
Quote:

Originally Posted by miguemate

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

And, therefore, it is sufficient to prove that $\displaystyle \lim_{x\to +\infty}|x|^{1/2}= +\infty$.
• Feb 10th 2010, 06:57 AM
CrazyCat87
Quote:

Originally Posted by HallsofIvy
And, therefore, it is sufficient to prove that $\displaystyle \lim_{x\to +\infty}|x|^{1/2}= +\infty$.

Why would that be sufficient?
• Feb 10th 2010, 07:52 AM
HallsofIvy
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 AM
CrazyCat87
ok I'm done proving it, thanks everyone for all your help!!