# Simple limit

1. Apr 27, 2010

### 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...

2. Apr 27, 2010

### Nick R

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

3. Apr 28, 2010

### HallsofIvy

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

4. May 6, 2010

### 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 $x\to\infty$.

5. May 9, 2010

Ts = op?

6. May 9, 2010

### Landau

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

7. May 9, 2010

### 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"}

8. May 9, 2010

### Mentallic

Hahaha that's a good one :rofl: