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Epsilon-delta proof of sqrt of abs value of x

  • #1
Prove that the limit as x->0 of [itex]\sqrt{\left|x\right|}[/itex] = 0

attempt:

[itex]\sqrt{\left|x\right|}[/itex] < [itex]\epsilon[/itex] when |x|<[itex]\delta[/itex]

-[itex]\epsilon[/itex] < [itex]\sqrt{\left|x\right|}[/itex] < [itex]\epsilon[/itex]

[itex]\left|x\right|[/itex] < [itex]\epsilon[/itex][itex]^{2}[/itex]

[itex]\delta[/itex]=[itex]\epsilon[/itex][itex]^{2}[/itex]

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?
 

Answers and Replies

  • #2
2,967
5
[tex]
\sqrt{|x|} < \epsilon \Leftrightarrow |x| < \epsilon^{2}
[/tex]

If you choose a [itex]\delta[/itex] such that:
[tex]
0 < \delta < \epsilon^2
[/tex]
you are done.
 
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