Question

If $f\left(x\right)$ is continuous at $x=0$, where $f\left(x\right)=\frac{(e^{3x}-1)\sin x}{x\log (x+1)}$ for $x\ne 0$. FInd $f\left(0\right).$

Answer 1

$f\left(x\right)=\frac{(e^{3x}-1)\sin x}{x\log (x+1)}$

$\Rightarrow f\left(0\right)={lim}_{x\to 0}f\left(x\right)$

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

This is in $\frac{0}{0}$ form, using L'ospital's rule, we get

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

This is in $\frac{0}{0}$ form, using L'ospital's rule, we get

$\Rightarrow f\left(0\right)={lim}_{x\to 0}\frac{-(e^{3x}-1)\sin x+\cos x\left(3e^{3x}\right)+\sin x\left(9e^{3x}\right)+3e^{3x}\cos x}{\frac{1}{(x+1{)}^2}+\frac{1}{x+1}}$

Substituting $x=0$ we get,

$\Rightarrow f\left(0\right)=\frac{3+3}{1+1}$

Therefore, $\Rightarrow f\left(0\right)=3$