$$\lim_{x\to 0^+} \sqrt{x} = 0$$
Proof:
The definition for a "right" limit is:
$\forall \epsilon > 0, \exists \delta_{\epsilon} > 0$ such that $\forall x \in \mathcal{D}$ (where $\mathcal{D}$ is the domain of the function) for which it is true that $0 < x – x_0 < \delta$ we have $|f(x) – \ell | < \epsilon$.
So we have
$$\sqrt{x}| < \epsilon$$
Now we know the square root outputs only positive numbers (in its principal branch, right?) so the absolute value is useless. Whence
$$x < \epsilon^2$$
At this point I can take $\delta = \epsilon^2$ such that we have $0 < x < \epsilon^2$ to prove $|\sqrt{x} – 0| < \epsilon$
Is this correct?
Now let's check for $x\to 0^-$ which doesn't exist. Following the similar definition (what changes for left limits is that we now have $0 < x_0 – x < \delta$):
$$|\sqrt{x} – 0 | < \epsilon$$
which implies that for $0 < x_0 – x < \delta$ I can prove $|\sqrt{x}| < \epsilon$. But the would mean
$$0 < -x < \delta \longrightarrow 0 > x > \delta$$
This is impossible for if $\delta = \epsilon^2$ as before, I cannot ever verify $0 > x > \delta$. So the limit doesn't exist.
Is this correct?
Eventually, how would I prove there is no limit as $x\to 0$ without specifying the direction?
Best Answer
Since $\sqrt x$ is undefined when $x<0$, it makes no sense to talk about $\lim_{x\to0^-}\sqrt x$. And it also follows that $\lim_{x\to0^+}\sqrt x=\lim_{x\to0}\sqrt x$, since both assertions $\lim_{x\to0^+}\sqrt x=0$ and $\lim_{x\to0}\sqrt x=0$ mean$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x>0\implies\left|\sqrt x\right|<\varepsilon.$$