Limit with variable: non-defined expression I have a given limit that depends on a variable $a$:
$$\lim_{x \rightarrow \infty} \left (\frac{e^{ax}}{1 - ax} \right)$$
I understand cases for $a < 0 \implies \lim = 0$ and $a > 0 \implies \lim = -\infty$. However, for the case $a = 0$, the expression $ax$ which is basically $0\cdot \infty$ in undefined. I somehow know, that the result will be $\lim (\frac{e^0}{1}) = 1$ but I am not sure how to justify that $0\cdot\infty$ is $0$ in this case.
Thanks for any ideas or an explanation!
 A: If we choose the value $a=0$ we evaluate the expression $\frac{e^{ax}}{1 - ax}$ before passing to the limit so your result would be
$$\lim_{x\to\infty}\left(\left[\frac{e^{ax}}{1 - ax}\right]_{a=0}\right)=\lim_{x\to\infty}\frac{e^0}{1-0}=1$$
A: There is no “indetermination” in $e^{ax}$ when $a=0$: it just means $e^0=1$, because $0x=0$.
You don't compute such a limit by plugging in $\infty$ in place of $x$, which wouldn't make sense. You can, however, use that


*

*$\lim_{x\to\infty}e^{ax}=\infty$ (for $a>0$);

*$\lim_{x\to\infty}e^{ax}=0$ (for $a<0$).


But this is different from simply plugging in $\infty$. For instance, in the case of $a<0$, you can conclude that
$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}=0
$$
because the numerator has $0$ limit and the denominator has $\infty$ limit.
On the other hand, you cannot immediately draw a conclusion in case $a>0$, since the numerator and the denominator have limit $\infty$ and $-\infty$, respectively. For this you can do a simple application of L’Hôpital's theorem:
$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}
\overset{(H)}{=}
\lim_{x\to\infty}\frac{ae^{ax}}{-a}
=
\lim_{x\to\infty}-e^{ax}=-\infty
$$
For the case $a=0$ you simply have
$$
\lim_{x\to\infty}\frac{e^{ax}}{1-ax}
=
\lim_{x\to\infty}\frac{1}{1}=1
$$
