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Could the following limit be computed without L'Hopital and Taylor? Thanks.

$\lim_{x\rightarrow0} \frac{\log(1+x)}{x^2}-\frac{1}{x}$

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    @Frank : $\log$ and $\ln$ are synonymous notations, except in contexts where logarithms to some base other than $e$ is expected. Normally in mathematics $\log$ with no subscript means $\log_e$.2012-06-11

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Here's an approach. Note that you can write the limit as

$\lim_{x\to 0} \frac{\log(1+x)-x}{x^2}$

and use the following definition:

$\log(1+x) = \int_1^{1+x}\frac{dt}{t}$

For $x$ small, the midpoint rule gives us that:

$\log(1+x) \approx \frac{x}{2} \left( 1 + \frac{1}{1+x}\right) = \frac{x+\frac{1}{2}x^2}{1+x}$

and hence we have

$\lim_{x\to 0} \frac{\frac{x+\frac{1}{2}x^2}{1+x}-x}{x^2} = \lim_{x\to 0}\frac{x+\frac{1}{2}x^2-x-x^2}{x^2} = \lim_{x\to 0}\frac{-\frac{1}{2}x^2}{x^2} = -\tfrac{1}{2}$

You may consider the midpoint rule to be a cheat, since it is typically justified using Taylor's theorem (I suspect that it can be proved without Taylor's theorem, though I haven't tried it).

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    @ChrisTaylor After reading your proof seriously, I find that your proof is not rigorously true. Although $A=\ln(1+x)\sim\frac12x(1+1/(1+x))=B$, we cannot immediately conclude that $f(A)\sim{}f(B)$ where $f(t)=(t-x)x^{-2}$.2012-06-11