Epsilon-delta proof of sqrt of abs value of x

1. Sep 26, 2011

kahwawashay1

Prove that the limit as x->0 of $\sqrt{\left|x\right|}$ = 0

attempt:

$\sqrt{\left|x\right|}$ < $\epsilon$ when |x|<$\delta$

-$\epsilon$ < $\sqrt{\left|x\right|}$ < $\epsilon$

$\left|x\right|$ < $\epsilon$$^{2}$

$\delta$=$\epsilon$$^{2}$

This seems to work...but I just ignored the negative epsilon since I cant square it in my inequality...I dont know if you are allowed to just ignore this?

2. Sep 26, 2011

Dickfore

$$\sqrt{|x|} < \epsilon \Leftrightarrow |x| < \epsilon^{2}$$

If you choose a $\delta$ such that:
$$0 < \delta < \epsilon^2$$
you are done.

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