<s>[INST] Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$ The limit is  
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
 [/INST] I assume you know about the geometric series and Taylor's theorem to integrate them, but you might not. I apologize if you don't, as this will not be of much use.
We have
$$
\frac 1{\log(x+1)} - \frac 1x = \frac{x-\log(x+1)}{x \log(x+1)} = \frac{1 - \frac{\log(x+1)}{x}}{\log(x+1)}.
$$
Now
$$
\frac 1{1+x} = \frac 1{1-(-x)} = \sum_{n \ge 0} (-x)^n, 
$$
hence
$$
\log(x+1) = \sum_{n \ge 0} \frac{-(-x)^{n+1}}{n+1} = \sum_{n \ge 0} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}2 + \frac{x^3}3 - \cdots.
$$
Also,
$$
\frac{\log(x+1)}{x} = \sum_{n \ge 0} \frac{(-1)^n x^n}{n+1},
$$
hence
$$
1 - \frac{\log(x+1)}x = \sum_{n \ge 1} \frac{(-1)^{n+1} x^n}{n+1} = \frac x2 - \frac{x^2}3 + \frac{x^3}4 - \cdots.
$$
Therefore,
$$
\lim_{x \to 0} \frac{1}{\log(x+1)} - \frac 1x = \lim_{x \to 0} \frac{\frac x2 - \frac{x^2}3 + \frac{x^3}4 - \cdots}{x - \frac{x^2}2 + \frac{x^3}3 - \cdots} = \lim_{x \to 0} \frac{\frac 12 - \frac{x}3 + \frac{x^2}4 - \cdots}{1 - \frac{x}2 + \frac{x^2}3 - \cdots} = \frac 12.
$$
Note : you don't need the whole series expansion, you could approximate to the second term using Taylor's theorem. But I guess that falls into your "big-O notation" category of proofs.
Hope that helps,
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