Question

Find the value of 'a' for which the function f defined as $f\left(x\right)=\left\{\begin{array}{c}a\sin \frac{\pi }{2}(x+1),x\le 0\\ \frac{\tan x-\sin x}{x^3},x>0\end{array}\right\}$ is continuous at $x=0$

Answer 1

$f\left(x\right)$ is continuous at $x=0$

$\Rightarrow$L.H.L of $f\left(x\right)$ at $x=0=$R.H.L of $f\left(x\right)$ at $x=0=f\left(0\right)$

$\Rightarrow limx\to 0^{-}f\left(x\right)=limx\to 0^{+}f\left(x\right)=f\left(0\right)$ ........$\left(1\right)$

Now, $\Rightarrow limx\to 0^{-}f\left(x\right)=limx\to 0^{-}a\sin \frac{\pi }{2}(x+1)$ since $f\left(x\right)=a\sin \frac{\pi }{2}(x+1)$ if $x\le 0$

$=limx\to 0^{-}a\sin (\frac{\pi }{2}+\frac{\pi x}{2})$

$=limx\to 0^{-}a\cos \frac{\pi x}{2}$

$=a\cos 0=a$

$\Rightarrow limx\to 0^{+}f\left(x\right)=limx\to 0^{+}\frac{\tan x-\sin x}{x^3}$ since $f\left(x\right)=\frac{\tan x-\sin x}{x^3}$ if $x>0$

$=limx\to 0^{+}\frac{\frac{\sin x}{\cos x}-\sin x}{x^3}$

$=limx\to 0^{+}\frac{\sin x-\sin x\cos x}{\cos x{x}^3}$

$=limx\to 0^{+}\frac{\sin x(1-\cos x)}{\cos x{x}^3}$

$=limx\to 0^{+}\frac{1}{\cos x}.limx\to 0^{+}\frac{\sin x}{x}\times \frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}\times 4}$ since $1-\cos x=2{\sin }^2\frac{x}{2}$

$=1\times 1\times \frac{1}{2}\times limx\to 0{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2$

$=\frac{1}{2}\times 1=\frac{1}{2}$

Also, $f\left(0\right)=a\sin \frac{\pi }{2}(0+1)=a\sin \frac{\pi }{2}=a$

Substituting these values in $\left(1\right)$ we get

$a=\frac{1}{2}$