Question

If $\displaystyle\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10, \alpha, \beta, \gamma \in R$ then the value of $\alpha+\beta+\gamma$ is ________.

Answer 1

$\displaystyle\lim _{x \rightarrow 0} \frac{\alpha x\left(1+x+\frac{x^{2}}{2}\right)-\beta\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}\right)+\gamma x^{2}(1-x)}{x^{3}}$
$\displaystyle\lim _{x \rightarrow 0} \frac{x(\alpha-\beta)+x^{2}\left(\alpha+\frac{\beta}{2}+\gamma\right)+x^{3}\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)}{x^{3}}=10$
For limit to exist
$\alpha-\beta=0, \alpha+\beta 2+\gamma=0$
$\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 \ldots \ldots $ (i)
$\beta=\alpha, \gamma=-3 \frac{\alpha}{2}$
Put in (i)
$\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10 $
$\frac{\alpha}{6}+\frac{3 \alpha}{2}=10 $
$\Rightarrow \frac{\alpha+9 \alpha}{6}=10$
$\Rightarrow \alpha=6 $
$\alpha=6, \beta=6, \gamma=-9$
$\alpha+\beta+\gamma=3$