well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with

L'Hopitale's rule...?

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- Mar 8th 2008, 05:22 PMakhayoonlimit

well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with

L'Hopitale's rule...? - Mar 8th 2008, 05:50 PMmr fantastic
Well, so you can use l'Hopital's rule.

To get the derivative of you can use logarithmic differentiation.

Edit: My mistake: , NOT e. l'H CANNOT be used. See post #4. - Mar 8th 2008, 05:59 PMakhayoon
still kind of shaky on this I tried multiplying everything by ln so I could reduce the

with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me - Mar 8th 2008, 06:08 PMmr fantastic
- Mar 8th 2008, 06:54 PMmr fantastic
The limit >is< meant to be x --> 0. so l'Hospital's Rule can be used.

To get the derivative of :

.

So the limit becomes:

.

Now show that .

Multiple applications of l'Hospital's rule is one possible approach.

Another approach is to substitute the series expansion for ln (1 + x): . This substitution is valid since the series converges for and ...... - Mar 9th 2008, 06:56 AMakhayoon
ok, so I evaluate the limit as it approaches zero to get -e/2 like u showed, but aren't I asked to evaluate it as it approaches infinity? so how does this cover that?

- Mar 9th 2008, 01:39 PMmr fantastic