Verification of Limit Using Taylor Expansion: x-ln(1+x)/x^2 = 1/2

[Fernando Revilla]

I quote a question from Yahoo! Answers

Use series to evaluate lim x->0 (x-ln(1+x)/x^2)?
I got 1/2. Could anyone please verify my answer. I am still very confused about series. Please show how you got your answer.
Thanks

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

Certainly, the limit is $1/2:$
$$\lim_{x\to 0}\frac{x-\log (1+x)}{x^2}=\lim_{x\to 0}\frac{x-\left(x-\frac{x^2}{2}+o(x^2)\right)}{x^2}\\
=\lim_{x\to 0}\left(\frac{1}{2}-\frac{o(x^2)}{x^2}\right)= \frac{1}{2}-0=\frac{1}{2}$$