Here is a function:
$$\frac{e^{ax}-e^x-x}{x^2}$$
In which $a$ is a coefficient. The problem whats us the value for $a$ which gives a finite value for the limit,
$$\lim_{x\to0}\frac{e^{ax}-e^x-x}{x^2}$$
How to find that value? I first though that the degree of the numerator must be equal to that of the denominator. but I don't know what the definition of degree is for $e^x$.
Even if I knew, I couldn't find the limit.
On
HINT:
As $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
$$e^{ax}-e^x-x=1-1+x(a-1-1)+x^2\cdot\dfrac{a^2-1}2+\text{terms containing higher powers of }x$$
On
Applying once L'Hôpital yields $$\lim_{x\to0}\frac{a e^{ax}-e^x-1}{2x},$$ which can be finite only if the numerator tends to $0$, i.e. $a=2$. Another application, with this assumption, shows that the limit is finite iff $a=2$.
On
While Appreciating others' answers, with the help of Olivier Oloa, I post my own solution.
$$e^{ax}=1+ax+\frac{a^2}{2}x^2$$
$$e^{x}=1+x+\frac{1}{2}x^2$$
$$\frac{e^{ax}-e^x-x}{x^2}=\frac{(a-2)+(\frac{a^2}{2}-1)x}{x}$$
We want a finite limit, so $\frac{a^2}{2}-1=1$ and $a-2=0$. Which both lead to $a=2$.
Evaluating the limit again with known $a$,
$$\lim_{x\to0}\frac{e^{2x}-e^x-x}{x^2}=\frac{1+2x+2x^2-1-x-\frac12x^2}{x^2}=\frac{(2-\frac12)x^2}{x^2}=1.5$$
Hint. One may recall that, as $u \to 0$, by the Taylor series expansion one has, $$ e^u=1+u+\frac{u^2}2+O(u^3). $$ Can you take it from here?