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

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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$\log(1+x)$ might be $\ln(1+x)$? – Frank Science Jun 11 '12 at 9:07
$\log$ is the preferred symbol for the Napier logarithm, here in Europe. – Siminore Jun 11 '12 at 9:08
@Frank Science: yes. – user 1618033 Jun 11 '12 at 9:52
@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 $\log_e$. – Michael Hardy Jun 11 '12 at 10:20

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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interesting approach. Thanks for your answer. – user 1618033 Jun 11 '12 at 9:24
See here en.wikipedia.org/wiki/Mean_value_theorem. It's what you mean by midpoint rule, which is used to prove Taylor's theorem, not proved by Taylor's theorem. – Frank Science Jun 11 '12 at 10:05
No, I definitely mean midpoint rule: en.wikipedia.org/wiki/Midpoint_rule – Chris Taylor Jun 11 '12 at 10:09
@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 $f(t)=(t-x)x^{-2}$. – Frank Science Jun 11 '12 at 11:52