Expert: Alon Mandes - 7/11/2008

Question
Hi,
I have a question similar to
lim x approaches infinity of e^-x times e^x.

It seems to me that the limit here would be 1, but the graph I have indicates the answer is 0.
How do I get started actually working this problem?

Answer
Ok, our function is e^(-xe^x). Let's get started first by
intuition: e^(-xe^x) can be written as [e^(-x)]^(e^x) which means(1/e^x)e^x. As we know, e^x grows rapidly, thus 1/e^x declines rapidly as well .(because it becomes fraction smaller than 1), & further more, "this fraction" is powered by e^x which itself grows rapidly. Now we conclude that our function will decline very very rapidly to zero. (Fraction)^x -->0 , (fraction)^(e^x)--->0 faster.

Now let's consider our claim analytically : Let's set y=e^(-xe^x), we
perform Ln for both sides, we get : Ln(y)=-xe^x. We can agree that
-xe^x goes to -(infinity) when x goes to infinity. Our question will be now : Which value of y can make the Ln function reaches the
-infinity ??? & the answer is 0. It's pretty obvious from the graph of the Ln function. In other words when x reaches infinity, y reaches zero. & y is exactly our function.

Summery : Lim (x-->inf) e^(-xe^x) = 0.

Alon.