Mathematics - Continuity - Differentiability Question with Solution | TestHub

For $f(x)$ to be differentiable at $x=0$, we must evaluate $f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}={\lim }_{h\to 0}\frac{h^a\sin (1/h^b)-0}{h}={\lim }_{h\to 0}{h}^{a-1}\sin (1/h^b)$. For this limit to exist, we need $a-1>0$, which implies $a>1$. In this case, $f^{\prime }(0)=0$. Now, for $x\ne 0$, the derivative $f^{\prime }(x)$ is given by $f^{\prime }(x)=\frac{d}{dx}\left(x^a\sin \left(\frac{1}{x^b}\right)\right)=a{x}^{a-1}\sin \left(\frac{1}{x^b}\right)+x^a\cos \left(\frac{1}{x^b}\right)\left(-\frac{b}{x^{b+1}}\right)=a{x}^{a-1}\sin \left(\frac{1}{x^b}\right)-b{x}^{a-b-1}\cos \left(\frac{1}{x^b}\right)$. For $f^{\prime }(x)$ to be continuous at $x=0$, we must have ${\lim }_{x\to 0}{f}^{\prime }(x)=f^{\prime }(0)=0$. This requires both terms in $f^{\prime }(x)$ to tend to 0 as $x\to 0$. For $a{x}^{a-1}\sin (1/x^b)\to 0$, we need $a-1>0$. For $-b{x}^{a-b-1}\cos (1/x^b)\to 0$, we need $a-b-1>0$. Combining these conditions, we need $a>1$ and $a>b+1$. The condition $a>b+1$ directly implies $a-b>1$. Therefore, the range of $a-b$ is $(1,\infty )$.