The value of a for which the function

Question:

The value of a for which the function

$f(x)=\left\{\begin{array}{ll}\frac{\left(4^{x}-1\right)^{3}}{\sin (x / a) \log \left\{\left(1+x^{2} / 3\right)\right\}}, & x \neq 0 \\ 12(\log 4)^{3} & , x=0\end{array}\right.$ may be continuous at $x=0$ is

(a) 1

(b) 2

(c) 3

(d) none of these

Solution:

(d) none of these

For $f(x)$ to be continuous at $x=0$, we must have

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

$\lim _{x \rightarrow 0}\left[\frac{\left(4^{x}-1\right)^{3}}{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}\right]=12(\log 4)^{3}$

$\Rightarrow \lim _{x \rightarrow 0}\left[\frac{\frac{\left(4^{x}-1\right)^{3}}{x^{3}}}{\frac{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}{x^{3}}}\right]=12(\log 4)^{3}$

$\Rightarrow \lim _{x \rightarrow 0}\left[\frac{a\left(\frac{4^{x-1}}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \frac{\log \left(1+\frac{x^{2}}{3}\right)}{x^{2}}}\right]=12(\log 4)^{3}$

$\Rightarrow 3 a \lim _{x \rightarrow 0}\left[\frac{\left(\frac{4^{x}-1}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}$

$\Rightarrow 3 a\left[\frac{\lim _{x \rightarrow 0}\left(\frac{4^{x}-1}{x}\right)^{3}}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \lim _{x \rightarrow 0} \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}$

$\Rightarrow 3 a(\log 4)^{3}=12(\log 4)^{3}$           $\left[\because \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)=\log a, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right.$ and $\left.\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$\Rightarrow a=4$