Mathematics - Limits Question with Solution | TestHub

We have,

$\underset{x\to 0}{\lim }\left[\frac{\alpha x{e}^x-\beta {\log }_e(1+x)+\gamma x^2{\mathrm{e}}^{-x}}{x{\sin }^{2}x}\right]=10$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{\alpha x\left(1+x+{\displaystyle \frac{x^2}{2!}+..}\right)-\beta \left(x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}-...\right)+\gamma x^2\left(1-x+{\displaystyle \frac{x^2}{2!}}-..\right)}{x^3\left({\displaystyle \frac{{\sin }^{2}x}{x^2}}\right)}\right]=10$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{\alpha x\left(1+x+{\displaystyle \frac{x^2}{2}}\right)-\beta \left(x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}\right)+\gamma x^2\left(1-x\right)}{x^3}\right]=10$

Neglecting higher powers.

$\underset{x\to 0}{\lim }\frac{x(\alpha -\beta )+x^2\left(\alpha +\frac{\beta }{2}+\gamma \right)+x^3\left(\frac{\alpha }{2}-\frac{\beta }{3}-\gamma \right)}{x^3}=10$

For limit to exist

$\alpha -\beta =0$

$\alpha +\frac{\beta }{2}+\gamma =0$

$\frac{\alpha }{2}-\frac{\beta }{3}-\gamma =10 \dots \left(\mathrm{i}\right)$

$\beta =\alpha ,\gamma =-\frac{3\alpha }{2}$

Putting in $\left(i\right)$, we get

$\frac{\alpha }{2}-\frac{\alpha }{3}+\frac{3\alpha }{2}=10$

$\Rightarrow \frac{\alpha }{6}+\frac{3\alpha }{2}=10$

$\Rightarrow \frac{\alpha +9\alpha }{6}=10$

$\Rightarrow \alpha =6$

Hence,

$\alpha =6,\beta =6,\gamma =-9$

$\Rightarrow \alpha +\beta +\gamma =3$