Question

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

Answer 1

Answer: 3

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