Indeterminate form : $ 0 \cdot \infty$ I have a question about the limit of :
$$\lim_{t \to \infty}t\cdot e^{-t}$$
Why is that equal to $0$ ? Shoudln't $ 0 \cdot \infty$ be an inderminate form ?
It's true that with the graphic we can determine that this limit is equal to  $0$ :

But all I know is that  : $ 0 \cdot \infty$ is supposed to be an inderminate form .
 A: $$\lim_{t\to \infty} te^{-t}=0$$
as your graph clearly shows. But: $$\int_0^\infty xe^{-x} dx =\lim_{t\to\infty}\int_0^t xe^{-x}dx=\lim_{t\to\infty}\bigg[1-e^{-t}(1+t)\bigg]=1$$
Your graph clearly shows the integral (area under the curve) is non-zero.
The answer to the broader question you pose about whether $0\cdot \infty$ is indeterminate is "Not necessarily".
Suppose $f(x)=x^2, g(x)=x$. Then $f(x)\to\infty, \frac{1}{g(x)}\to 0$, but $\frac{f(x)}{g(x)}=x\to\infty$
Swap the two functions around and we see $\frac{g(x)}{f(x)}=\frac 1x \to 0$
What the function converges to just depends on the rate of convergence of each side.
A: $ 0.\infty $ is of course an indeterminate form but in this particular case, it is zero.
Put $ t=\ln(u)$.
then
$$te^{-t}=\ln(u)e^{\ln(\frac 1u)}=\frac{\ln(u)}{u}$$
when $ t\to +\infty $, $ u\to +\infty$, and, as you know, $\frac{\ln(u)}{u}\to 0$
because, for $ u $ great enough, it is easy to prove that $ \ln(u)<\sqrt{u}.$
A: A nice way to show this is:
$$f(t)=te^{-t}=\frac{t}{e^t}$$
and we know that:
$$e^t=\sum_{n=0}^\infty\frac{t^n}{n!}=1+t+\frac{t^2}{2}+\frac{t^3}{6}+...$$
and so:
$$f(t)=\frac{t}{1+t+t^2/2+t^3/3+...}=\frac{1}{1/t+1+t/2+t^2/3+...}$$
now if we look at this we can see that:
$$\lim_{t\to\infty}\frac1t=0$$
and so on the bottom we have something of the form:
$$\frac 1{1+\infty+\infty+...}$$
which will obviously tend to $0$
