# Homework Help: L'hospital's Rule

1. Nov 28, 2007

### redsox5

1. The problem statement, all variables and given/known data

lim$$_{x -> infin}$$ x[(x+1)ln(1+(1/x))-1]

3. The attempt at a solution

Am I able to start this with the chain rule. I'm not sure how to start this.

2. Nov 28, 2007

### rocomath

i've been simplifying your original problem to see what form of L'Hospital's you have

$$\lim_{x \rightarrow \infty} x[(x+1)\ln(1+\frac{1}{x})-1]$$

also is this correct?

3. Nov 28, 2007

### redsox5

well that's the oringal problem
Is it an infinity times infinity problem, but to use l'hospital's rule it has to be a fraction

4. Nov 28, 2007

### redsox5

can i leave x on top and move [(x+1)ln(1+(1/x))-1] to the bottom by raising it to the -1 power? Then use l'hospital's rule

5. Nov 28, 2007

### nrqed

I don't know if this is going to be useful to you because I don;t know if you have seen that trick but I can get the answer quickly by using a Taylor expansion of $$ln(1+\epsilon) \approx 1 + \epsilon$$ for small epsilon. Then simple algebra leads dircetly to the answer.

6. Nov 28, 2007

### redsox5

ok think i may have a solution

lim x-> infin $$\frac{x}{[(x+1)ln(1+(1/x))-1]^-1}$$

then use l'hospital's rule and get

$$\frac{1}{-[(x+1)ln(1+(1/x))-1]^-1 * [ln(1+(1/x)+(x+1)(1/1+(1/x))}$$

7. Nov 28, 2007

### redsox5

yea, i'm not reallu sure about that. I'm not sure if it will be acceptable

8. Nov 28, 2007

### Dick

Be careful here, you may be quickly lead to the wrong answer. You need to keep the quadratic term in the expansion as well. I got burnt by that once.

9. Nov 28, 2007

### redsox5

you mean in my use of the chain rule?..is this the right approach though?

10. Nov 28, 2007

### Dick

This limit isn't even the same as the one you started out with. Write it as
[(x+1)*ln(1+1/x)-1]/(1/x). Now it's 0/0 (though that's not completely clear until you show (x+1)*ln(1+1/x) -> 1. You can use l'Hopital but you've got to be careful.

11. Nov 28, 2007

### redsox5

so after you use l'hospital's rule the first time you get

$$\frac{ln(1+\frac{1}{x})+(x+1)(\frac{1}{1+\frac{1}{x}})}{\frac{-1}{x^{2}}}$$

and that's infin/0 ...right?

12. Nov 28, 2007

### Dick

You were the one who brought up the chain rule. The derivative of ln(1+1/x) is [1/(1+1/x)]*(-1/x^2). It's 0/0. Do some algebra on that before you differentiate again.

13. Nov 28, 2007

### redsox5

wait..i the correct answer -1 since it equals infin/neg infin

14. Nov 28, 2007

### Dick

infin/neg infin is NOT -1. It's indeterminant. You need to differentiate again. In the end you will find a perfectly finite limit.

15. Nov 28, 2007

### redsox5

how many time do you have to use l'hospital's rule..it's getting really messy

16. Nov 28, 2007

### redsox5

after the second time i use it i have 1/0

17. Nov 28, 2007

### Dick

This is the last time. But clean up the algebra in the numerator before you differentiate again. Better to do it before than after.

18. Nov 28, 2007

### redsox5

but it's 1/0 can I still use l'hospitals'?

19. Nov 28, 2007

### Dick

I get [ln(1+1/x)-1/x]/(-1/x^2) after I do the algebra from the first differentiation. That's 0/0, not 1/0.

20. Nov 28, 2007

### redsox5

your numerator is wrong, her's the oringal again

lim x-> infin x[(x+1)ln(1+(1/x))-1]