$\text{ If }L=\underset{x\to 0}{lim} \frac{e^{ax}-e^x-x}{x^2}\text{ is a finite real number, then }$

$\begin{array}{l}\text{ Given }L=\underset{x\to 0}{lim} \frac{e^{ax}-e^x-x}{x^2}=\text{ finite }\left(\frac{0}{0}\text{ form }\right)\\ \text{ Using L'Hospitals rule }\\ L=\underset{x\to 0}{lim} \frac{e^{ax}a-e^x-1}{2x}\dots \dots \dots \dots \dots ...\left(1\right)\end{array}$

$\begin{array}{l}\left(\begin{array}{c}\because \frac{a-1-1}{2\left(0\right)}=\text{ finite }\\ \Rightarrow \frac{a-2}{0}=\text{ finite }\end{array}\right)\\ \therefore a-2=0\Rightarrow a=2\\ \text{ Substituting }a=2\text{ in (1) }\\ L=\underset{x\to 0}{lim} \frac{2e^{2x}-e^x-1}{2x}\left(\frac{0}{0}\text{ form }\right)\end{array}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{4e^{2x}-e^x}{2}\\ =\frac{4-1}{2}\\ =\frac{3}{2}\\ \therefore a=2,L=\frac{3}{2}\end{array}$