Epsilon-delta proof of sqrt of abs value of x

[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 can't square it in my inequality...I don't know if you are allowed to just ignore this?

 

[Dickfore]

<br /> \sqrt{|x|} &lt; \epsilon \Leftrightarrow |x| &lt; \epsilon^{2}<br />

If you choose a \delta such that:
<br /> 0 &lt; \delta &lt; \epsilon^2<br />
you are done.