# Thread: Limit from the right

1. ## Limit from the right

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?

2. (Assuming the limit is to 0+). Um, let $\displaystyle \epsilon > 0$. We must find a $\displaystyle \delta > 0$, s.t. if $\displaystyle 0<x<\delta$, $\displaystyle \sqrt{x}<\epsilon$. Since $\displaystyle \sqrt{\;\cdot\;}$ is increasing, $\displaystyle \sqrt{x} < \sqrt{\delta}$. So chuse $\displaystyle \delta \le \epsilon^2$ and it should follow...

Note that delta is a function of epsilon only. You can get amusing results if it is allowed to depend on x. http://math.berkeley.edu/~critch/Pi_equals_0,_or,_how_I_learned_to_stop_worrying_an d_let_delta_depend_on_x.pdf