Answer
$a$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as: $e^{ax}=1+ax+\dfrac{(ax)^2}{2!}+\dfrac{(ax)^3}{3!}+.......$
Now, we have:
$\lim\limits_{x \to 0} \dfrac{e^{ax}-1}{x}=\dfrac{(1+ax+\dfrac{(ax)^2}{2!}+\dfrac{(ax)^3}{3!}+.......)-1}{x}\\=\lim\limits_{x \to 0} (a+\dfrac{a^2x}{2!}+\dfrac{a^3x^2}{3!}+.........)\\=a$
Therefore, we have: $\lim\limits_{x \to 0} \dfrac{e^{ax}-1}{x}=a$
