Answer

Solution

1. 1

   Using $e^{a_1 x}-1 = a_1 x + \tfrac{(a_1 x)^2}{2}+o(x^2)$, we get

2. 2
   $$\frac{e^{a_1 x}-1}{a_2 x^{2}+a_3 x} = \frac{a_1 x + O(x^2)}{a_3 x + a_2 x^2} = \frac{a_1 + O(x)}{a_3 + a_2 x}\xrightarrow[x\to 0]{}\frac{a_1}{a_3}.$$
3. 3

   For the limit to exist and equal $1$, we must have $a_3\ne 0$ and $a_1=a_3$. (If $a_3=0$, the denominator behaves like $a_2 x^2$, and the limit would diverge or be zero unless $a_1=0$, in which case the limit is $0\ne 1$.) Thus the necessary condition is $a_1=a_3\ne 0$.

Answer structure / marking notes

Content note