Question

$f\left(x\right)=\{\begin{array}{c}\frac{{(3\sin x-1)}^2}{x\log (1+x)},x\ne 0\\ 2\log 3,x=0\end{array}$

Check continuity at $x=0$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{{(3\sin x-1)}^2}{x\log (1+x)},x\ne 0\\ 2\log 3,x=0\end{array}$

Checking continuity at $x=0$

$\Rightarrow {lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}\frac{{\left[3\sin (0+h)\right]}^2}{(0+h)\log (1+0+h)}={lim}_{h\to 0}\frac{2(3\sin h-1)\left(3\cos h\right)}{1\log (1+h)+\frac{h}{1+h}}$

$={lim}_{h\to 0}\frac{2(3\sin h-1)\left(3\cos h\right)}{\frac{1}{1+h}+\frac{1}{1+h}+h{\left(\frac{1}{1+h}\right)}^2}$

$=9\to 1$

$\Rightarrow {lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}\frac{{(3\sin (0-h)-1)}^2}{(0-h)\log (1-h)}={lim}_{h\to 0}\frac{2(-3\sin h-1)(-3\cos h)}{-1\log (1-h)-\frac{1}{1+h}}$

$={lim}_{h\to 0}\frac{6\left(3\cos h\right)}{\frac{1}{1-h}+\frac{1}{1-h}-0}$

$=9\to 2$

$\Rightarrow f\left(0\right)=2\log \left(3\right)\to 3$

From $1,2\&3,f\left(x\right)$

For making it continuous, $f\left(0\right)=e^{2\log 3}=9$