# Solve this

Question:

If $f(x)=\left\{\begin{array}{ll}a \sin \frac{\pi}{2}(x+1), & x \leq 0 \\ \frac{\tan x-\sin x}{x^{3}}, & x>0\end{array}\right.$ is continuous at $x=0$, then a equals

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(C) $\frac{1}{4}$

(d) $\frac{1}{6}$

Solution:

(a) $\frac{1}{2}$

Given: $f(x)=\left\{\begin{array}{l}a \sin \frac{\pi}{2}(x+1), x \leq 0 \\ \frac{\tan x-\sin x}{x^{3}}, x>0\end{array}\right.$

We have

(LHL at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} a \sin \left(\frac{\pi}{2}(-h+1)\right)=a \sin \left(\frac{\pi}{2}\right)=a$

(RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} \frac{\tan h-\sin h}{h^{3}}$

$=\lim _{h \rightarrow 0} \frac{\frac{\sin h}{\cos h}-\sin h}{h^{3}}$

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

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

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2} \tan h}{4 \times \frac{h^{2}}{4} \times h}$

$=\frac{2}{4} \lim _{h \rightarrow 0} \frac{\sin ^{2} \frac{h}{2} \tan h}{\frac{h^{2}}{4} \times h}$

$=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \times \lim _{h \rightarrow 0} \frac{\tan h}{h}$

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

$=\frac{1}{2}$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

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