Question

If $\underset{x \rightarrow 0}{\text{Lim}} \frac{a \sin x +b x e^{x}+3 x^{2}}{\sin x-2 x+\tan x}$ exists and has value equal to $L$, then the value of $\frac{b-L}{a}$, is equal to

Answer 1

$L=\underset{x \rightarrow 0} {\text{Lim}}\frac{a\left(x-\frac{x^{3}}{3 !}+\ldots\right)+b x\left(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots\right)+3 x^{2}}{\left(\frac{\sin x-x}{x^{3}}-\frac{(x-\tan x)}{x^{3}}\right) x^{3}}$
$L=6 \underset{x \rightarrow 0} {\text{Lim}} \frac{x(a+ b)+x^{2}(b+3)+x^{3}\left(\frac{b}{2}-\frac{a}{6}\right)+\ldots}{x^{3}}$
$\Rightarrow L=6\left(\frac{b}{2}-\frac{a}{6}\right)$
$\Rightarrow a+b=0$ and $b+3=0$
$\Rightarrow b=-3$. Also $a=3$
$L=6\left(-\frac{3}{2}-\frac{3}{6}\right)=-12$
Hence, $\left(\frac{b-L}{a}\right)=\frac{-3-(-12)}{3}=\frac{9}{3}=3$