Rigorously Evaluating the Limit of $\tmop{te}^{- t}$

[Nick R]

Hello, it is pretty obvious that the following limit is equal to zero:

$Lim t \rightarrow \infty (\tmop{te}^{- t}) = 0$

For example, for t=100 it is 100*e^{-100}

But how would you take this limit "rigorously"? I tried decomposing the function with a mclaurin series and te^-t is equal to this series:

$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1} t^n}{(n - 1) !}$

How would I actually evaluate this series for t->infinity? Or is this the wrong approach?

Also for a finite number of terms it appears that this series diverges...

 

[Nick R]

Errrr... L'hopital's rule. Sorry should have spent a while longer thinking about it before posting.

 

[HallsofIvy]

Write it as t/e^t and use L'Hopital's rule as Nick R suggested.

 

[Landau]

Nick R = TS ;)

A more direct proof: since e^x = 1 + x + x^2+ ..., it is obvious that e^x>x for all x\in\mathbb{R}. In other words, \frac{e^x}{x}>1. Hence

\frac{e^x}{x}=\frac{1}{2}\left(\frac{e^{x/2}}{x/2}\right)e^{x/2}>\frac{1}{2}e^{x/2}\to\infty if x\to\infty.

It follows that xe^{-x}=\frac{x}{e^x}\to 0 if x\to\infty.

 

[The Chaz]

Ts = op?

 

[Landau]

I'm sorry, with TS I meant Topic (/Thread) Starter. Is OP (original poster?) more standard?

 

[The Chaz]

We are mathematicians. We can call it whatever we want! But it is mandatory to use at least two of these:
1) greek letter(s)
2) subscript
3) AlTeRnAtInG CaPs

I recommend that we define \tau\sigma_{1}(399107):= {"Nick R"}

 

[Mentallic]

We are mathematicians. We can call it whatever we want! But it is mandatory to use at least two of these:
1) greek letter(s)
2) subscript
3) AlTeRnAtInG CaPs

I recommend that we define \tau\sigma_{1}(399107):= {"Nick R"}

Hahaha that's a good one