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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!

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  • $\begingroup$ For $a=0$, your expression becomes $\frac{1}{1-x}$, which most certainly goes to 0 (from the left) as $x \to \infty$. Do you maybe mean to let $a \to 0$ and $x \to \infty$ at the same time? $\endgroup$ – fgp May 11 '13 at 12:54
  • $\begingroup$ @fgp I mean the situation where $a = 0$ AND $x \rightarrow \infty$ $\endgroup$ – Smajl May 11 '13 at 12:59
  • $\begingroup$ Then you need to update your question to reflect that. You also have to specify how $a$ goes to $0$ while $x$ goes to $\infty$. The whole point of this excersize is to demonstrate that the limit can depend on that... $\endgroup$ – fgp May 11 '13 at 13:15
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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 $$\lim_{x\to\infty}\left(\left[\frac{e^{ax}}{1 - ax}\right]_{a=0}\right)=\lim_{x\to\infty}\frac{e^0}{1-0}=1$$

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  • $\begingroup$ It didnt occur to me that it works like this.. thanks! $\endgroup$ – Smajl May 11 '13 at 13:22
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 May 11 '13 at 13:25
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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

  1. $\lim_{x\to\infty}e^{ax}=\infty$ (for $a>0$);
  2. $\lim_{x\to\infty}e^{ax}=0$ (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 $$ \lim_{x\to\infty}\frac{e^{ax}}{1-ax} = \lim_{x\to\infty}\frac{1}{1}=1 $$

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