Question

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

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}a\sin \frac{\pi }{2}(x+1),& x\le 0\\ \frac{\tan x-\sin x}{x^3},x>0& \end{array}$

LHL at $x=0$,

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f(-h)$

$\Rightarrow \underset{h\to 0}{lim}a\sin \frac{\pi }{2}(-h+1)=a\sin \frac{\pi }{2}=a$

RHL at $x=0$,

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}f\left(h\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{\tan h-\sin h}{h^3}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin h}{\cos h}-\sin h}{h^3}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin h}{\cos h}(1-\cos h)}{h^3}$

$=\underset{h\to 0}{lim}\frac{\tan h(1-\cos h)}{h^3}$

$=\underset{h\to 0}{lim}\frac{2{\sin }^2\frac{h}{2}\tan h}{4\frac{h^2}{4}\times h}$

$=\frac{2}{4}\underset{h\to 0}{lim}\frac{{\sin }^2\frac{h}{2}\tan h}{\frac{h^2}{4}\times h}$

$=\frac{1}{2}\underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\cdot \underset{h\to 0}{lim}\frac{\tan h}{h}$

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

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

Therefore,

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

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