Mathematics - Continuity - Differentiability Question with Solution | TestHub

For continuity at $x=0$:${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}\frac{x}{1+e^{1/x}}$. As $x\to 0^{+}$, $1/x\to \infty$, so $e^{1/x}\to \infty$. We can write this as ${\lim }_{x\to 0^{+}}\frac{x{e}^{-1/x}}{e^{-1/x}+1}=\frac{0}{0+1}=0$. ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}\frac{x}{1+e^{1/x}}$. As $x\to 0^{-}$, $1/x\to -\infty$, so $e^{1/x}\to 0$. Thus, ${\lim }_{x\to 0^{-}}\frac{x}{1+e^{1/x}}=\frac{0}{1+0}=0$. Since ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{+}}f(x)=f(0)=0$, $f(x)$ is continuous at $x=0$. For differentiability at $x=0$:$f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}={\lim }_{h\to 0}\frac{\frac{h}{1+e^{1/h}}-0}{h}={\lim }_{h\to 0}\frac{1}{1+e^{1/h}}$. $f^{\prime }(0^{+})={\lim }_{h\to 0^{+}}\frac{1}{1+e^{1/h}}=\frac{1}{1+\infty }=0$. $f^{\prime }(0^{-})={\lim }_{h\to 0^{-}}\frac{1}{1+e^{1/h}}=\frac{1}{1+0}=1$. Since $f^{\prime }(0^{+})\ne f^{\prime }(0^{-})$, $f(x)$ is not differentiable at $x=0$. Therefore, $f(x)$ is continuous but not differentiable at $x=0$.