Answer (18)

Solution

Finding right hand limit

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

$=\lim _{h \rightarrow 0} f(h)$

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

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

Let $\cos ^{-1}\left(1-h^{2}\right)=\theta \Rightarrow \cos \theta=1-h^{2}$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}}$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^{2}}}}$

$=\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}}$

$R=\frac{\pi}{\sqrt{2}}$

Now finding left hand limit

$$ \begin{aligned} & L=\lim _{x \rightarrow 0^{-}} f(x) \\ & =\lim _{h \rightarrow 0} f(-h) \\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-{-h}^{2}\right) \sin ^{-1}(1-{-h})}{{-h}-{-h}^{3}} \\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^{2}\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^{3}} \\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^{2}+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^{2}\right)} \\ & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^{2}\right)} \\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^{2}+2 h}\right) \\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \end{aligned} $$