Re: Help on limit with log

$\displaystyle \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} - \cdots$

$\displaystyle \ln(1+x)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} - \cdots$

$\displaystyle \frac{\ln(1+x)-x}{x^2}=-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+\frac{x^3}{5} - \cdots$

Easier to see now?

Re: Help on limit with log

Wow!, now it makes sense. Thank you very much :).

Re: Help on limit with log

That would work if you were allowed to use series. But that is equivalent to L'Hopital's rule":

$\displaystyle \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f(0)+f'(0)x+c_1x^2+\dots}{g(0)+g'(0)x+c_2x^2 +\dots} $

and assuming $\displaystyle f(0)=g(0)=0$,

$\displaystyle \lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f'(0)+c_1x+\dots}{g'(0)+c_2x+\dots} = \frac{f'(0)}{g'(0)}$

- Hollywood

Re: Help on limit with log

Thanks for the reply.

I am allowed to use L'hopital's rule. But I am also asked to do it with a different method.

Given that in another exercise I had to expand ln(1+x) using a taylor series (which gives as a result the series abender posted) I think that is the extra way they want me to do it.

Re: Help on limit with log

I see. You're right - abender's calculation is undoubtedly what they are looking for.

- Hollywood