The value of a so that $\underset{x\to 0}{lim} \frac{1}{x^2}\left(e^{\alpha x}-e^x-x\right)=\frac{3}{2}$ is

We have,

$\begin{array}{l}\underset{x\to 0}{lim} \frac{\left(e^{\alpha x}-e^x-x\right)}{x^2}=\frac{3}{2}\\ \Rightarrow \underset{x\to 0}{lim} \frac{\left\{1+\alpha x+\frac{(\alpha x{)}^2}{2!}+\dots \right\}-\left\{1+x+\frac{x^2}{2!}+\dots \right\}-x}{x^2}=\frac{3}{2}\end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\underset{x\to 0}{lim} \frac{(\alpha -2)x+\left({\alpha }^2-1\right){\displaystyle \frac{x^2}{2!}}+\frac{x^3}{3!}\left({\alpha }^3-1\right)+\dots }{x^2}=\frac{3}{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha -2=0\Rightarrow \alpha =2\\ \end{array}$