Existence of limit of given function

Question

If a function $f:\mathbb{R} \to \mathbb{R}$ is defined as $f(x) = \sqrt{|x|}$, then does the limit $\displaystyle \lim_{x \to y} \frac{f(x) - f(y)}{x - y}$ exist for each $y \in \mathbb{R}$?

My approach: By rationalising, we have that \begin{align} \lim_{x \to y} \frac{f(x) - f(y)}{x - y} &= \lim_{x \to y} \frac{\sqrt{|x|} - \sqrt{|y|}}{x - y}\\ &= \lim_{x \to y} \frac{|x| - |y|}{(x - y)\left(\sqrt{|x|} + \sqrt{|y|}\right)}\\ &= \left(\lim_{x \to y} \frac{1}{\sqrt{|x|} + \sqrt{|y|}}\right)\cdot \lim_{x \to y} \frac{|x| - |y|}{x - y}\\ &= \frac{1}{2\sqrt{y}}\lim_{x \to y} \frac{|x| - |y|}{x - y} \end{align} Now the numerator and the denominator of the other limit approaches 0, so I can use L'Hopital's. However, I do not want to use L'Hopital's, because I'm studying from an analysis text that hasn't yet introduced the derivative. All in all though, my expectation is that the limit does exist. But is there a better/more elegant way to show this?

Answer 1

Since the questions asks whether the limit exists for all $y\in\mathbb{R}$, it suffices to show that it does not exist for some value of $y$. In particular, we show that the limit does not exist when $y=0$.

In this case we have the limit as

$$\lim_{x\to0}\frac{\sqrt{\lvert x\rvert}}{x}.$$

If we consider the two one-sided limits, then

$$\lim_{x\downarrow0}\frac{\sqrt{\lvert x\rvert}}{x}=\lim_{x\downarrow0}\frac{\sqrt{x}}{x}=\lim_{x\downarrow0}\frac{1}{\sqrt{x}}=\infty,$$

$$\lim_{x\uparrow0}\frac{\sqrt{\lvert x\rvert}}{x}=\lim_{x\downarrow0}\frac{\sqrt{\lvert -x\rvert}}{-x}=\lim_{x\downarrow0}\left(-\frac{1}{\sqrt{x}}\right)=-\infty,$$

and so the limit does not exist.

Answer 2

For the case where $y>0$, when $x$ goes to $y$ you can always have $x>0$ eventually. Then $f$ is $\sqrt{x}$ and it's easy, the case when $y<0$ is nearly the same. But for $y=0$ you will find the limit different when $x$ goes to $0$ being positive or negative (+-infty), therefore the limit exist for $y\in\mathbb{R}-\{0\}$.