Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^{\alpha }\cos \left(\frac{1}{x}\right),if& x\ne 0\\ 0,if& x=0\end{array}is$ continuous at $x=0$ then

• A. $\alpha <0$
• B. $\alpha >0$
• C. $\alpha =0$
• D. $\alpha \ge 0$

Answer 1

The correct option is B $\alpha >0$

$f\left(x\right)=\{\begin{array}{c}{x}^{\alpha }C.D\left(\frac{1}{x}\right),ifx\ne 0\\ 0,ifx=0\end{array}$

continuous at $x=0$

Left hand limit

$L.H.L=\underset{x\to 0^{-}}{lim}f\left(x\right)=(0^{-}{)}^{\alpha }C.D\left(\frac{1}{0^{-}}\right)$

We know that $\frac{1}{0^{-}}\Rightarrow -\infty$

But $C.D(-\infty )\to$ always will lie $(-1,1]$

It means $C.D\left(\frac{1}{0^{-}}\right)$ will be finite quantity.

Similarly It $I$ check Right hand limit then-

$R.H.L\underset{x\to 0^{+}}{lim}=(0^{+}{)}^{\alpha }C.D\left(\frac{1}{0^{+}}\right)$

Again same $\frac{1}{0^{+}}\Rightarrow \infty$

$C.D\left(\frac{1}{0^{+}}\right)\Rightarrow$ will lie $(-1,1)$

Now is $\alpha <0\Rightarrow \left(0^{+}{)}^{negative}\right(-1,1)\Rightarrow \frac{1}{(0^{+}{)}^{positive}}(-1,1)$

$\Rightarrow$ this will give undefine

So, is $\alpha >0$ then $\left(0^{+}{)}^{positive}\right(-1,1)=0(-1,1)=0$

Similarly $\left(0^{-}{)}^{positive}\right(-1,1)=0[-1,1]=0$

Ans it $\alpha =0$, then $L.H.L$ will lie $[-1,1]$

and $R.H.L$ will lie $[-1,1]$

So, for this we can not say clearly that this the limit value

So, for $\alpha >0\to$ function is continuous at $x=0$