Question

Show that the function $f\left(x\right)=\{\begin{array}{cc}{x}^2,& x\le 1\\ \frac{1}{x},& x>1\end{array}$ is continuous at $x=1$ but not differentiable.

Answer 1

To be continuous at $x=1$, $li{m}_{x\to 1^{-}}f\left(x\right)=li{m}_{x\to 1^{+}}f\left(x\right)=f\left(x\right)$
Now for $x\le 1$ $f\left(x\right)=x^2$ and $x^2$ is continuous so $li{m}_{x\to 1^{-}}f\left(x\right)=f\left(x\right)$

Now we need to check $li{m}_{x\to 1^{-}}f\left(x\right)=li{m}_{x\to 1^{+}}f\left(x\right)$

So at $li{m}_{x\to 1^{-}}f\left(x\right)=li{m}_{x\to 1^{-}}{x}^2=(1{)}^2=1$

and at ${lim}_{x\to 1^{+}}f\left(x\right)={lim}_{x\to 1^{+}}\frac{1}{x}=\frac{1}{1}=1$

So, $li{m}_{x\to 1^{-}}f\left(x\right)=li{m}_{x\to 1^{+}}f\left(x\right)$.

Hence $f\left(x\right)$ is continuous at $x=1$.

To be differentiable at $x=1$, $f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$

for $x\le 1$ $f\left(x\right)=x^2\to f^{\prime }\left(x\right)=2x\to f^{\prime }\left(1^{-}\right)=2\left(1\right)=2$

for $x>1$ $f\left(x\right)=\frac{1}{x}\to f^{\prime }\left(x\right)=-\frac{1}{x^2}\to f^{\prime }\left(1^{+}\right)=-\frac{1}{1^2}=1$

So $f^{\prime }\left(1^{-}\right)\ne f^{\prime }\left(1^{+}\right)$ means $f\left(x\right)$ not differentiable at $x=1$.