Question

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

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 $x=0=f\left(0\right)$

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

Now,${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}a\sin \frac{\pi }{2}(x+1)$

$\because f\left(x\right)=a\sin \frac{\pi }{2}(x+1)$ if $x\le 0$

$={lim}_{x\to 0}a\sin (\frac{\pi }{2}+\frac{\pi }{2}x)$

$={lim}_{x\to 0}a\cos \frac{\pi }{2}x=a\cos 0=a$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{\tan x-\sin x}{x^3}$

$\because f\left(x\right)=\frac{\tan x-\sin x}{x^3}$ if $x>0$

$={lim}_{x\to 0}\frac{\frac{\sin x}{\cos x}-\sin x}{x^3}$

$={lim}_{x\to 0}\frac{\sin x-\sin x\cos x}{\cos x.x^3}$

$={lim}_{x\to 0}\frac{1}{\cos x}{lim}_{x\to 0}\frac{\sin x}{x}{lim}_{x\to 0}\frac{1-\cos x}{x^2}$

$={lim}_{x\to 0}1\times 1\times {lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}\times 4}$

$={lim}_{x\to 0}1\times 1\times \frac{1}{2}=\frac{1}{2}$

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

Putting above values in $\left(1\right)$ we get,$a=\frac{1}{2}$