# L'hospital's Rule

## Homework Statement

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

## The Attempt at a Solution

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

## Answers and Replies

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?

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

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

nrqed
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## Homework Statement

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

## The Attempt at a Solution

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

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.

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

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

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

Dick
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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.

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.

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

Dick
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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

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

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.

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?

Dick
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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.

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

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

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

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

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

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

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

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

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

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.

your numerator is wrong, her's the oringal again

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

Gib Z
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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.

Tiny correction, there is no constant term in that. So its just epsilon, though as Dick said, we should keep the quadratic term in there as well.

Dick
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your numerator is wrong, her's the oringal again

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

I don't need the original. That's at the top of the page. I would like to know what you think the result of the first application of l'Hopital's rule is.

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