Practicing Success
Let $f(x)=\left\{\begin{array}{cc}(x-1) \sin \left(\frac{1}{x-1}\right), & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.$ Then, which one of the following is true? |
f is differentiable at x = 0 but not at x = 1 f is differentiable at x = 1 but not at x = 0 f is neither differentiable at x = 0 nor at x = 1 f is differentiable at x = 0 and at x = 1 |
f is differentiable at x = 0 but not at x = 1 |
We observe that $\lim\limits_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ $=\lim\limits_{x \rightarrow 1} \frac{(x-1) \sin \left(\frac{1}{x-1}\right)}{x-1}=\lim\limits_{x \rightarrow 1} \sin \left(\frac{1}{x-1}\right)$ = An oscillating number between -1 and 1. ∴ $\lim\limits_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ does not exist. ⇒ f(x) is not differentiable at x = 1. and, $\lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ $=\lim\limits_{x \rightarrow 0} \frac{(x-1) \sin \left(\frac{1}{x-1}\right)-\sin 1}{x}$ $= \lim\limits_{x \rightarrow 0} \frac{x \sin \left(\frac{1}{x-1}\right)}{x}-\lim\limits_{x \rightarrow 0} \frac{\sin \left(\frac{1}{x-1}\right)+\sin 1}{x}$ $= -\sin 1-\lim\limits_{x \rightarrow 0} \frac{2 \sin \frac{x}{2(x-1)} \cos \left\{\frac{2-x}{2(x-1)}\right\}}{\left\{\frac{x}{2(x-1)}\right\} 2(x-1)}$ = - sin 1 + cos 1 ⇒ f(x) is differentiable at x = 0. |