1. ## Limits

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

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

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

$lim_{x\to 0^+} f(x) =+\infty$ 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

4. 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 $\lim_{x\to +\infty}|x|^{1/2}= +\infty$.

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

6. Because making the substitution y= -x in $\lim_{x\to -\infty} |x|^{-1/2}$ gives $\lim_{y\to\infty} |-y|^{-1/2}= \lim_{y\to\infty} |y|^{-1/2}$ which is, of course, the same as $\lim_{x\to \infty}|x|^{-1/2}$

7. ok I'm done proving it, thanks everyone for all your help!!