What have I done wrong? Calculating $\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$ 
I want to calculate $$\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$$
using L'Hospitals rule:
$$\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}\overbrace{=}^{L'Hospital}\lim\limits_{x\to 0}\frac{\frac{2}{1+2x}}{2x}=\lim\limits_{x\to 0}\frac{4x}{2x+1}\to \frac{0}{1}=0$$

The solution from my lecture is that $\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$ doesn't exist. 
We could show this with $\lim\limits_{x\to 0^+}\frac{\ln(1+2x)}{x^2}\neq\lim\limits_{x\to 0^-}\frac{\ln(1+2x)}{x^2}$ (I did this in another post).

But how do I exactly show, that $\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$ does not exist while applying L'Hospitals rule or is there no way one can show that with L'Hospital? What am I missing? What did I do wrong while applying the rule?
 A: Note that
$$\dfrac{\dfrac{2}{1+2x}}{2x}=\dfrac{2}{2x(1+2x)}.$$
Now
$$\lim_{x\to 0^-}\dfrac{2}{2x(1+2x)}=-\infty$$ and
$$\lim_{x\to 0^+}\dfrac{2}{2x(1+2x)}=+\infty.$$
A: Your second equality is not correct. Consider $\frac{1}{2}$. Divide $\frac{1}{2}$ by 2. It is not 1, but rather 1/4. Because
$\frac{1}{2}$ $\div$ 2 $=$ $\frac{1}{2}$ $(\frac{1}{2})$ $=$ $\frac{1}{4}$
A: We have 
\begin{align}
\lim\limits_{x\to 0^+}\frac{\ln(1+2x)}{x^2}
=&
\lim\limits_{x\to 0^+}\frac{2}{x}\cdot\ln(1+2x)^{\frac{1}{2x}}
\\
=&\lim\limits_{x\to 0^+}\frac{2}{x}\cdot \lim\limits_{x\to 0^+}\ln(1+2x)^{\frac{1}{2x}}
\\
=&
\\
=&\lim\limits_{x\to 0^+}\frac{2}{x}\cdot \ln \lim\limits_{x\to 0^+}(1+2x)^{\frac{1}{2x}}
\\
=&\lim\limits_{x\to 0^+}\frac{2}{x}\cdot \ln e
\\
=&\lim\limits_{x\to 0^+}\frac{2}{x}\cdot 1
\\
=&\infty
\end{align}
and 
\begin{align}
\lim\limits_{x\to 0^-}\frac{\ln(1+2x)}{x^2}
=&
\lim\limits_{x\to 0^-}\frac{2}{x}\cdot\ln(1+2x)^{\frac{1}{2x}}
\\
=&\lim\limits_{x\to 0^-}\frac{2}{x}\cdot \lim\limits_{x\to 0^-}\ln(1+2x)^{\frac{1}{2x}}
\\
=&
\\
=&\lim\limits_{x\to 0^-}\frac{2}{x}\cdot \ln \lim\limits_{x\to 0^-}(1+2x)^{\frac{1}{2x}}
\\
=&\lim\limits_{x\to 0^-}\frac{2}{x}\cdot \ln e
\\
=&-\infty
\end{align}
