Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?

Question

Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?

I thought that if there exists a limit on both sides then it would be differentiable?

Answer 1

Here's the graph of $f:=\sqrt{\vert x-2 \vert}$ where $-5 \leq x \leq 5$:

Here $f$ is a continuous curve but there is a "corner" at $x=2$, so $f$ is not differentiable there.

For the formal argument, the limit $$\lim_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim_{h \rightarrow 0} \frac{\sqrt{\vert 2+h-2 \vert} \;-\sqrt{\vert 2-2 \vert} }{h}=\lim_{h \rightarrow 0} \frac{\sqrt{\vert h \vert}}{h} $$ does't exist!

Answer 2

Since

$$\lim_{h\to0^+}\frac{\sqrt{|2+h-2|}-\sqrt{|2-2|}}{h}=\lim_{h\to0^+}\frac{1}{\sqrt{|h|}}$$

does not exist, $\sqrt{|x-2|}$ is not differentiable at $x=2$. (Althogh both $\displaystyle \lim_{x\to 0^+}\sqrt{|x-2|}$ and $\displaystyle \lim_{x\to 0^-}\sqrt{|x-2|}$ exist.)

Answer 3

Functions can be continuous without being differentiable. This is an example. If the function has a corner it isn't differentiable at that point. A simpler example is $f(x)=|x|$ around $x=0$. At any point less than $0$ the derivative is $-1$. At any point greater than $0$ the derivative is $1$. If you try to apply the definition of the derivative at $0$ you will fail. We have $$f'(0)=\left.\frac {df(x)}{dx}\right|_{x=0}=\lim_{h \to 0}\frac {f(h)-f(0)}h=\begin {cases}-1&x \lt 0\\1 & x \gt 0 \end {cases}$$ and there is no limit at $0$ so $f$ is not differentiable there.

Answer 4

If a function is differentiable, then it implies that it is continuous. However, the opposite is not true.

For example: |x| is not differentiable at x=0, although it's continuous. Also for this function $\sqrt|x-2|$, the right hand limit at x=2 and left hand limit at x=2 does not match, so limit doesn't exist at x=2, hence not differentiable.

$$\\ \textbf{LHL: } \lim_{h\rightarrow 2^{-}} \frac{ f(2+h)-f(2)}{h} = \lim_{h\rightarrow 2^{-}} \frac{ \sqrt (2+h-2)-\sqrt(2-2)}{h}\\ =\lim_{h\rightarrow 2^{-}} \frac{ \sqrt (h)-\sqrt(0)}{h}\\ =\lim_{h\rightarrow 2^{-}} \frac{ \sqrt (h)}{h} = \frac{1}{\sqrt 2} \\ \textbf{RHL: } \lim_{h\rightarrow 2^{+}} \frac{ f(2+h)-f(2)}{h} = \lim_{h\rightarrow 2^{+}} \frac{ \sqrt (2-h-2)-\sqrt(2-2)}{h}\\ =\lim_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)-\sqrt(0)}{h}\\ =\lim_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)}{h} = \text{not defined}\\ \\\\ \because \textbf{RHL} \neq \textbf{LHL} $$ This is not a differentiable at x=2

Another simple way to check for differentiability is that there should not be a sharp corner in the curve.

In the plot of the $\sqrt|x-2|$, this corner occurs at $ x=2 $ (as shown in the figure) plot here