# Limit from the right

• April 8th 2010, 12:01 PM
sfspitfire23
Limit from the right
Hey guys,

Show that $\lim_{x \to a^+} \sqrt{x} = 0$ with a delta epsilon proof.

I first took $|\sqrt{x}-0| \rightarrow |\sqrt{x}|\times (|\sqrt{x}-0|) \rightarrow |x-0|< |\sqrt{x}| \times \epsilon$. So, $\delta = |\sqrt{x}| \times\epsilon$...is this right? How do I get the result to follow?
• April 8th 2010, 02:20 PM
(Assuming the limit is to 0+). Um, let $\epsilon > 0$. We must find a $\delta > 0$, s.t. if $0, $\sqrt{x}<\epsilon$. Since $\sqrt{\;\cdot\;}$ is increasing, $\sqrt{x} < \sqrt{\delta}$. So chuse $\delta \le \epsilon^2$ and it should follow...