Question

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

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

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

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

Answer 1

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


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

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

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

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\sin h}{\cos h}}-\sin h}{h^3} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\sin h}{\cos h}}\left({\displaystyle 1}-\cos h\right)}{h^3} \\ =\underset{h\to 0}{\lim }\frac{\left(1-\cos h\right)\tan h}{h^3} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}\tan h}{4\times {\displaystyle \frac{h^2}{4}}\times h} \\ =\frac{2}{4}\underset{h\to 0}{\lim }\frac{{\sin }^2{\displaystyle \frac{h}{2}}\tan h}{{\displaystyle \frac{h^2}{4}}\times h} \\ =\frac{1}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\right)}^2\times \underset{h\to 0}{\lim }\frac{\tan h}{h} \\ =\frac{1}{2}\times 1\times 1 \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{If}f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0,\mathrm{then} \\ \underset{\mathrm{x}\to 0^{-}}{\lim }f\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0^{+}}{\lim }f\left(\mathrm{x}\right) \\ \Rightarrow a=\frac{1}{2} \end{gathered}$$