Discussion Forum

1)

If the function f(x) defined by

$f(x)=\begin{cases}x \sin\frac{1}{x} ,&for  x \neq 0\\k, & for x=0 \end{cases}$

 is continuous at x=0 , then k is equal to

Answer:

Option A

Explanation:

Given 

$f(x)=\begin{cases}x \sin\frac{1}{x} ,&for  x \neq 0\\k, & for x=0 \end{cases}$

 iSince , f(x) is continuous at x=0

, $\therefore$    LHL=RHL=f(0)

 .......(i)

 Now, LHL$=\lim_{x \rightarrow 0-}f(x)=\lim_{h \rightarrow 0}f(0-h)$

   $=\lim_{h \rightarrow 0}(0-h)\sin \frac{1}{(0-h)}=\lim_{h \rightarrow 0}-h\sin\left(-\frac{1}{h}\right)$

 $=\lim_{h \rightarrow 0}h \sin \frac{1}{h}=0\times$ finite value =0

                                         $\left[\because \lim_{h \rightarrow 0}\sin \frac{1}{h}=finite  value\right]$

 and f(0)=k

$\therefore$   From Eq.(i)   , LHL= RHL

$\Rightarrow$    0=k

$\Rightarrow$    k=0