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Limit using Taylor expansion

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
Certainly, the limit is $1/2:$
$$\lim_{x\to 0}\frac{x-\log (1+x)}{x^2}=\lim_{x\to 0}\frac{x-\left(x-\frac{x^2}{2}+o(x^2)\right)}{x^2}\\
=\lim_{x\to 0}\left(\frac{1}{2}-\frac{o(x^2)}{x^2}\right)= \frac{1}{2}-0=\frac{1}{2}$$