Show that $f( x) =\sqrt[5]{x+2}$ has no derivative at $x=-2$.

Solution First we note that $f(-2) =\sqrt[5]{-2+2}=0.$ Then we investigate the limit. $$\lim\limits_{x\rightarrow -2}\dfrac{f( x) -f(-2) }{ x-(-2) }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}-0}{x+2 }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}}{x+2}=\lim\limits_{x \rightarrow -2}\dfrac{1}{\sqrt[5]{( x+2) ^{4}}}=\infty$$

Since the limit is infinite, the derivative of $f$ does not exist at $-2$.