<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] It is bit difficult to avoid LHR or series expansions here. I present here a technique which is almost like using series expansion, but a bit simpler conceptually. For this purpose I need to use the standard definition of $\log x$ as $\int_{1}^{x}(1/t)\,dt$.
Let us assume that $0 < t < 1$. Then it can be checked using algebra that $$1 - t < \frac{1}{1 + t} < 1 - t + t^{2}$$ If $0 < x < 1$ then upon integrating above inequality in the interval $[0, x]$ we get $$x - \frac{x^{2}}{2} < \log (1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$ or $$\dfrac{1}{x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3}} < \dfrac{1}{\log(1 + x)} < \dfrac{1}{x - \dfrac{x^{2}}{2}}$$ or $$\frac{6}{6x - 3x^{2} + 2x^{3}} < \frac{1}{\log(1 + x)} < \frac{2}{2x - x^{2}}$$ Subtracting $(1/x)$ from each term in above inequality we get (after some simplification) $$\frac{3 - 2x}{6 - 3x + 2x^{2}} < \frac{1}{\log(1 + x)} - \frac{1}{x} < \frac{1}{2 - x}$$ Taking limits as $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1}{\log(1 + x)} - \frac{1}{x} = \frac{1}{2}$$ To handle the case when $x \to 0^{-}$ we need to substitute $x = -y$ to get $$\frac{1}{\log(1 + x)} - \frac{1}{x} = \frac{1}{\log(1 - y)} + \frac{1}{y}$$ and $y \to 0^{+}$.
Next we can see that if $0 < y < 1$ then $$\log(1 - y) = \log(1 - y^{2}) - \log(1 + y)$$ Using $\log(1 - y^{2}) < -y^{2}$ and $\log(1 + y) > y - (y^{2}/2)$ we can see that $$\log(1 - y) < -y - \frac{y^{2}}{2}$$ or $$\log(1 - y) + y < -\frac{y^{2}}{2}\,\,\,\cdots (1)$$ Again we can see that $$\frac{y^{2}}{y^{2} - 1} < \log(1 - y^{2})$$ and $$\log (1 + y) < y - \frac{y^{2}}{2} + \frac{y^{3}}{3}$$ so that $$\log(1 - y) > \frac{y^{2}}{y^{2} - 1} - y + \frac{y^{2}}{2} - \frac{y^{3}}{3}$$ or $$\frac{y^{2}}{y^{2} - 1} + \frac{y^{2}}{2} - \frac{y^{3}}{3} < \log(1 - y) + y \,\,\,\cdots (2)$$ From the equations $(1)$ and $(2)$ we can see that $$\frac{1}{y^{2} - 1} + \frac{1}{2} - \frac{y}{3} < \frac{\log(1 - y) + y}{y^{2}} < -\frac{1}{2}$$ Taking limits as $y \to 0^{+}$ and using Squeeze theorem we get $$\lim_{y \to 0^{+}}\frac{\log(1 - y) + y}{y^{2}} = -\frac{1}{2}$$ It is now easy to observe that
$\displaystyle \begin{aligned}\lim_{y \to 0^{+}}\frac{1}{\log(1 - y)} + \frac{1}{y} &= \lim_{y \to 0^{+}}\frac{\log(1 - y) + y}{y\log(1 - y)}\\
&= \lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{-y^{2}\cdot\dfrac{\log(1 - y)}{-y}}\\
&= \lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{-y^{2}\cdot 1}\\
&= -\lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{y^{2}}\\
&= \frac{1}{2}\end{aligned}$
The above derivation is bit lengthy because it establishes the inequalities satisfied by $\log (1 + x)$ function using integration and their extensions to negative values of $x = -y$ by further algebraic manipulation. This method is the conceptually simpler (but taking more space and calculations) equivalent of using the Taylor's expansion $\log(1 + x) = x - x^{2}/2 + x^{3}/3 - \cdots$ In my view it is better to use the Taylor's expansion or LHR for such problems. However even when we apply Taylor or LHR it is better to change the problem into a different form as follows:
$\displaystyle \begin{aligned}\lim_{x \to 0}\frac{1}{\log(1 + x)} - \frac{1}{x} &= \lim_{x \to 0}\frac{x - \log(1 + x)}{x\log(1 + x)}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}\cdot\dfrac{\log(1 + x)}{x}}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}\cdot 1}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}}\\\end{aligned}$
Doing this above simplification avoids taking reciprocal of a series (if you use Taylor's expansion) and also avoids complicated differentiation (if you use LHR).
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