Question

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

Answer 1

Answer: 3

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