$\epsilon$-$\delta$ proof for $\lim_{x\to 0^+}\frac{1}{\sqrt{x}}$ does not exist.

Question

$\epsilon$-$\delta$ proof for $\lim\limits_{x\to 0^+} \dfrac{1}{\sqrt{x}}$ does not exist.

I can explain limit goes infinity, but how can show that, let $\lim\limits_{x\to 0^+} \dfrac{1}{\sqrt{x}}=L$ and find a argument for L does not exist?

Answer 1

Suppose $$\lim_{x\to 0^+}\frac{1}{\sqrt{x}} = L < \infty$$ so that for every $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |x| < \delta$ implies $\left|\frac{1}{\sqrt x} - L\right| < \epsilon$. Fix an $\epsilon > 0$ and the corresponding $\delta > 0$.

Now, $|x| < \delta$ implies $\frac{1}{\sqrt x} > \frac{1}{\sqrt{\delta}}$. By the triangle inequality, for all $x$ satisfying $|x| < \delta$ we have $$\frac{1}{\sqrt{x}} \le \left|\frac{1}{\sqrt x} - L\right| + |L| \le |L| + \epsilon < \infty$$ To find a contradiction, we shall pick some $x_0 > 0$ such that $x_0 < \delta$ but $\frac{1}{\sqrt{x_0}} > |L| + \epsilon$. The choice $$x_0 = \frac{1}{2} \min \left\{\delta, \frac{1}{(|L| + \epsilon)^2} \right\}$$ works. I leave it to you to see how.