Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{(4^x-1{)}^3}{\sin \left(\frac{x}{a}\right)\ln (1+\frac{x^2}{3})},& x\ne 0\\ 9(\ln 4{)}^3,& x=0\end{array}$ is continuous at $x=0$, then the value of $a$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

For $f\left(x\right)$ to be continuous at $x=0$, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)\cdots \left(1\right)$
$\Rightarrow \underset{x\to 0}{lim}\frac{(4^x-1{)}^3}{\sin \left(\frac{x}{a}\right)\ln (1+\frac{x^2}{3})}=\underset{x\to 0}{lim}\frac{(4^x-1{)}^3}{x^3}\cdot \frac{\left(\frac{x}{a}\right)}{\sin \left(\frac{x}{a}\right)}\cdot \frac{\frac{x^2}{3}}{\ln (1+\frac{x^2}{3})}\cdot 3a$
$=3a\cdot {\left(\underset{x\to 0}{lim}\frac{4^x-1}{x}\right)}^3\cdot \left(\frac{1}{\underset{x\to 0}{lim}\frac{\sin \left(x/a\right)}{\left(x/a\right)}}\right)\cdot \left(\frac{1}{\underset{x\to 0}{lim}\frac{\ln (1+x^2/3)}{\left(x^2/3\right)}}\right)$
$=3a\cdot (\ln 4{)}^3\cdot \frac{1}{1}\cdot \frac{1}{1}=3a(\ln 4{)}^3$
Using $\left(1\right),$ we get
$\Rightarrow 9(\ln 4{)}^3=3a(\ln 4{)}^3\therefore a=3$