# A simple limit that causes a lot of headache

1. Mar 31, 2009

### Marin

Hi everyone!

consider the following limit:

$$\lim_{x\rightarrow\infty}\frac{ln(1-\frac{1}{x})}{e^{-x}}$$

Since we get [0/0] (by injectivity of both exp and log functions), it smelled to me like de l'Hospital's rule until I began calculating the derivatives. Then I realised it's somehow useless...

Besides the correct value of the limit, I am also very interested in the point, why the rule is not applicable here :)

I appreciate every idea or hint you give me :)

marin

2. Mar 31, 2009

### CompuChip

The rule applies, I think, but it doesn't help you as it keeps giving 0/0.
You could consider the derivative and some careful estimates show that it is always negative (and exponentially grows in the -infinity direction).
Or you could use a formal argument after rewriting something like
$$\lim_{x \to \infty} \ln(1 - 1/x) e^{-x} = \lim_{y \to 0} \ln(1 - y) / e^y = e^{\ln(\ln(1 - y))} e^{-y}$$

3. Mar 31, 2009

### Marin

ok, i'll stick to the formal argument. Could you please elaborate on what exactly you do there. It's obviously not the substitution y=1/x, is it, since you should've changed the whole argument of the exp funct. and not just replace its sign? Well, how does it work then?

4. Mar 31, 2009

### CompuChip

Whoops, my mistake. You are right, I cannot do that.

It is intuitively clear that e^{-x} will go to 0 faster than ln(1 - 1/x) will. Perhaps substituting y = 1/x and making a Taylor expansion around y = 0?

Also note that the derivative argument, if written down correctly, is no less formal.

5. Mar 31, 2009

### Ben Niehoff

As $x \to \infty$, 1/x is small, so the logarithm can be approximated by

$$\ln (1 - \frac1x) \approx - \frac1x$$

Then it should be clear what is happening in the limit.

ETA:

Actually, l'Hopital's should have worked just fine. I get

$$-\lim_{x \to \infty} \frac{e^x}{x-x^2}$$

6. Mar 31, 2009

### JG89

Remember that $$ln(1 - 1/x) = - \int_{1 - 1/x}^1 \frac{1}{u} du$$

So your expression is $$-e^x \int_{1 - 1/x}^1 \frac{1}{u} du$$. By the MVT of integral calculus $$\int_{1 - 1/x}^1 \frac{1}{u} du = \frac{1}{x} \frac{1}{c}$$ where c is some value between 1 - 1/x and 1.

Your expression in your limit is now $$\frac{-e^x}{xc}$$. Since c is some value between 1 - 1/x and 1, we see that as x tends to infinity 1/c tends to 1 and it is easy to see that the rest of the expression then goes to negative infinity.

7. Mar 31, 2009

Double Post

8. Apr 1, 2009

### Marin

Ben Niehoff, you're right, l'Hospital works fine here after doing a bit of algebra :), btw, I think you've got a minus sign more/less in your answer, since the answer is -infty :)

JG89, I find your solution using the MVT very elegant, thanks for posting it :)

CompuChip, Taylor expansion also works very well :)

So, thanks once again for the help!