Help on limit with log

**bizan** · December 30th 2012, 06:53 AM

Hello, I have spent several hours stuck on this limit:

$lim(x->0) \frac{ln(1+x)-x}{x^2}$

I managed to solve it by L'hopital, its limit is -0.5 but I was wondering if anyone can come up with a different method, as I am not supposed to use L'hopital at this stage.

I tried

$lim\hspace{0.1cm} ln(1+x)^{\frac{1}{x^2} } - lim \frac{1}{x}$ = $ln \hspace{0.1cm} lim(1+x)^{\frac{1 1}{x x}} - lim \frac{1}{x}$ = $ln e^{1/x} - lim \frac{1}{x}$

But it did not work.

Any idea? Thanks!

**abender** · December 30th 2012, 07:36 AM

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

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

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

Easier to see now?

**bizan** · December 30th 2012, 07:40 AM

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

.

**hollywood** · December 30th 2012, 05:31 PM

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

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

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

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

- Hollywood

**bizan** · December 30th 2012, 05:50 PM

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.

**hollywood** · December 30th 2012, 06:58 PM

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

- Hollywood

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