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.
 A: $$\lim_{x\to0}\frac1{\ln(x+1)}-\frac1x=\lim_{x\to1}\frac1{\ln(x)}-\frac1{x-1}=L$$
since $x\to1$ is equivalent to $x^2\to1$, we can write
$$L=\lim_{x\to1}\frac1{\ln(x^2)}-\frac1{x^2-1}=\lim_{x\to1}\frac1{2\ln(x)}+\frac12\left(\frac1{x+1}-\frac1{x-1}\right)=\frac12L+\lim_{x\to1}\frac12\frac1{x+1}=\frac12L+\frac14$$
Hence, $L=\frac12L+\frac14$ and $L=\frac12$. 
A: 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,
A: You can use a series expansion to calculate the limit. The Maclaurin series (verify this for yourself) of 
$$\frac{1}{\ln(x+1)} = \frac{1}{x} +\frac{1}{2} - \frac{x}{12} +\frac{x^2}{24} + \frac{19x^3}{720} + \cdots$$
Now place that series expansion in your limit and you see that all the limit will be $\frac{1}{2}$
A: 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).
