The value of a , so that _ x → 0 (1)/(x2)(e^ ax - e^ x - x) =...

We have,

$\lim_{x \to 0}\dfrac{1}{x2}\left(e^{ax} - e^{x} - x\right) = \dfrac{3}{2}$

$⇒ \lim_{x \to 0}\left(\dfrac{ae^{ax} - e^{x} - 1}{2x}\right) = \dfrac{3}{2}$ (using $L$'Hospital rule to the left hand side)

As $x \to 0$, the denominator is zero& and as the value of the limit is finite so, numerator must be zero for $x \to 0$.

Hence, $a - 1 - 1 = 0 ⇒ a = 2$.