Thank you for reporting, we will resolve it shortly
Q.
If $\displaystyle\lim _{x \rightarrow 0} \frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x \sin ^2 x}=\frac{2}{3}$,
where $\alpha, \beta, \gamma \in R$, then which of the following is NOT correct ?
$\displaystyle\lim _{x \rightarrow 0} \frac{\alpha\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\right)+\beta\left(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\ldots\right)+\gamma\left(x-\frac{x^3}{3 !}+\ldots\right)}{\left.x^3\right)}$
constant terms should be zero
$\Rightarrow a+\beta=0$
coeff of $x$ should be zero
$\Rightarrow \alpha-\beta+\gamma=0$
coeff of $x^2$ should be zero
$\displaystyle\lim _{x \rightarrow 0} \frac{x^3\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)+x^4\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)}{x^3}=\frac{2}{3}$
$ \Rightarrow \frac{\alpha}{2}+\frac{\beta}{2}=0 $
$ \frac{\alpha}{6}-\frac{\beta}{6}-\frac{\gamma}{6}=2 / 3$
$ \Rightarrow \alpha=1, \beta=-1, \gamma=-2$