Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\tan x-\sin x}{x^3};& x<0\\ \frac{{\cot }^{-1}x-{\cos }^{-1}x}{x^3};& x>0\\ \frac{1}{2};& x=0\end{array}$

Then which of the following is correct

• A. $f\left(x\right)$ is continuous at $x=0$
• B. Both $\underset{x\to 0^{-}}{lim}f\left(x\right)$ and $\underset{x\to 0^{+}}{lim}f\left(x\right)$ exist, but $\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)$
• C. Both $\underset{x\to 0^{-}}{lim}f\left(x\right)$ and $\underset{x\to 0^{+}}{lim}f\left(x\right)$ exist, and $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne f\left(0\right)$
• D. Limits does not exist

Answer 1

The correct option is A $f\left(x\right)$ is continuous at $x=0$

$LHL=\underset{x\to 0^{-}}{lim}\frac{\sin x-\sin x\times \cos x}{x^3\cos x}=\underset{x\to 0^{-}}{lim}\frac{\sin x(1-\cos x)}{x^3\cos x}=\underset{x\to 0^{-}}{lim}\frac{\sin x}{x}\times \underset{x\to 0^{-}}{lim}\frac{2{\sin }^2\frac{x}{2}}{(\frac{x}{2}{)}^2\times 4}\times \underset{x\to 0^{-}}{lim}\frac{1}{\cos x}=1\times \frac{2}{4}\times \frac{1}{1}=\frac{1}{2}$

$RHL=\underset{x\to 0^{+}}{lim}\frac{{\cot }^{-1}x-{\cos }^{-1}x}{x^3}\left[\frac{0}{0}\text{form}\right]=\underset{x\to 0^{+}}{lim}\frac{\frac{-1}{1+x^2}+\frac{1}{\sqrt{1-x^2}}}{3x^2}=\underset{x\to 0^{+}}{lim}\frac{(1+x^2{)}^2-(1-x^2)}{3x^2\left(\sqrt{1-x^2}\right)(1+x^2)}\times \frac{1}{(1+x^2)+\sqrt{1-x^2}}=\underset{x\to 0^{+}}{lim}\frac{x^2+3}{3\left(\sqrt{1-x^2}\right)(1+x^2)}\times \frac{1}{(1+x^2)+\sqrt{1-x^2}}=\frac{3}{3\times 1\times 1}\times \frac{1}{1+1}=\frac{1}{2}$

$f\left(a\right)=f\left(0\right)=\frac{1}{2}$
$\Rightarrow LHL=RHL=f\left(a\right)\therefore f\left(x\right)\text{is continuous at}x=0$