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

Question

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

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

Answer 1

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).