# Simple way to get a complex Limit?

1. Mar 11, 2008

### BackEMF

I'm helping a few people out with school maths and I came across this limit (in connection with the Ratio Test):

$$\lim_{n \to \infty}\left(\frac{n}{n+1}\right)^n$$

now I can figure out the answer to this myself using l'Hôpital's Rule but the student's I'm teaching wouldn't have come across it yet, so it's either a misprint in their book - or there's a simple way of figuring it out which I can't see. Anyone have any ideas?

2. Mar 11, 2008

### grmnsplx

could we look at the limit of $$(\left(\frac{n}{n+1}\right)^n - \left(\frac{(n-1)}{(n-1)+1}\right)^{n-1} )$$ ?

I mean if $$\lim_{n \to \infty} (a_{n} - a_{n-1}) \to \infty$$ then doesn't that tell us something?

Last edited: Mar 11, 2008
3. Mar 11, 2008

### BackEMF

Sorry, I probably shouldn't have mentioned the ratio test at all, I'm just looking to find the limit, which equals $$e^{-1}$$ which I got by taking the natural log of the term above and applying l'Hôpital's Rule.

4. Mar 11, 2008

### grmnsplx

you want to use the squeeze theorem

5. Mar 11, 2008

### BackEMF

OK, any ideas of what two functions to pick?

6. Mar 11, 2008

### Mute

Surely they have come across the definition of $e$? If so, it's simple enough:

$$\left(\frac{n}{n+1}\right)^n = \left(\frac{n+1}{n}\right)^{-n} = \left[\left(1 + \frac{1}{n}\right)^n\right]^{-1}$$

Taking the limit of the last expression and using the definition of $e$ gives $e^{-1}$.

Is there a reason why they wouldn't be allowed to do it this way?

7. Mar 12, 2008

### grmnsplx

mute is correct

8. Mar 12, 2008

### BackEMF

Excellent, yes they would have come across that definition of $e$. But as for myself on the other hand...let's just say it's been a while.

Thanks Mute and grmnsplx.