Hey guys,

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

I first took $\displaystyle |\sqrt{x}-0| \rightarrow |\sqrt{x}|\times (|\sqrt{x}-0|) \rightarrow |x-0|< |\sqrt{x}| \times \epsilon$. So, $\displaystyle \delta = |\sqrt{x}| \times\epsilon$...is this right? How do I get the result to follow?