Question

If $f\left(x\right)=\{\begin{array}{cc}x\sin \left(1/x\right)& forx\ne 0\\ 0& forx=0\end{array}$ then

• A. $f$ is a continuous function
• B. $f^{\prime }(0+)$ exists but $f^{\prime }(0-)$ does not exist
• C. $f^{\prime }(0+)=f^{\prime }(0-)$
• D. $f^{\prime }(0+)$ and $f^{\prime }(0-)$ do not exist

Answer 1

Answer: (A), (D)

$f$ is a continuous function
$f^{\prime }(0+)$ and $f^{\prime }(0-)$ do not exist

$\underset{x\to 0}{lim}xsin\left(\frac{1}{x}\right)$
$=\underset{x\to 0}{lim}\frac{sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$
$=0$
And $f\left(0\right)=0$.
Hence f(x) is continuous at x=0.
Now
$f^{\prime }\left(x\right)=sin\left(\frac{1}{x}\right)-xcos\left(\frac{1}{x}\right).\frac{1}{x^2}$
$=sin\left(\frac{1}{x}\right)-\frac{cos\frac{1}{x}}{x}$
Now
$\underset{x\to 0}{lim}\frac{cos\frac{1}{x}}{x}$, doesn't exists.
Hence
$f^{\prime }\left(0^{+}\right)$ and $f^{\prime }\left(0^{-}\right)$ does not exists.