Solving $\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}$ without L'Hospital or Sandwiching Using the L'Hospital I got: $\lim\limits_{x  \to 0} \frac{\ln(x+1)}{\ln(1+4x+x^2)} = \lim\limits_{x  \to 0}\frac{\frac{1}{1+x}}{\frac{4+2x}{1+4x+x^2}}= \frac{1}{4}$,
I then wondered, if I could sandwich it andobserved that for: $\frac{x-1}{x} \le \ln(x) \le x-1 $, that:$ \frac{1}{4} \xleftarrow{x \to 0} \frac{\frac{x+1-1}{x+1 }}{1+4x+x^2-1} \le \frac{\ln(x+1)}{\ln(1+4x+x^2)} \le \frac{x+1-1}{\frac{1+4x+x^2-1}{1+4x+x^2}} \xrightarrow{x \to 0}\frac{1}{4} $
I then however wondered, if it were directly possible to get the result by cleverly manipulating the variables, so that the term simplifies itself to $\frac{1}{4}$ in the limit. Simply substituting $x+1$ or $1+4x+x^2$ with $e^u$ seems rather cumbersone, if (!) it is constructive at all. I am afraid, that there is just no real nice simplification of above. But maybe someone knows another nice, elegant way to get this limit :) 
I am always happy to expand and practice my toolkit. 
As always thanks in advance. 

Note that we can write $\log(1+4x+x^2)=\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)$.
Therefore, we have
$$\begin{align}
\frac{\log(1+x)}{\log(1+4x+x^2)}&=\frac{\log(1+x)}{\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)}\\\\&=\frac{\log(1+x)}{\log(1+4x)}\left(1+\frac{\log\left(1+\frac{x^2}{1+4x}\right)}{\log(1+4x)}\right)^{-1}
\end{align}$$
Inasmuch as $\displaystyle \lim_{x\to0}\left(1+\frac{\log\left(1+\frac{x^2}{1+4x}\right)}{\log(1+4x)}\right)=1$, the problem boils down to evaluating the limit
$$\begin{align}
\lim_{x\to 0}\frac{\log(1+x)}{\log(1+4x)}&=\frac14\lim_{x\to 0}\left(\frac{\log(1+x)}{x}\,\,\frac{4x}{\log(1+4x)}\right)\\\\
&=\frac14
\end{align}$$
where we used  
$$\begin{align}
\lim_{h\to 0}\frac{\log(1+h)}{h}&=\lim_{h\to 0}\frac{\log(1+h)-\log(1)}{h}\\\\
&=\left. \left(\frac{d\log(x)}{dx}\right)\right|_{x=1}\\\\
&=\left. \left(\frac{1}{x}\right)\right|_{x=1}\\\\
&=1
\end{align}$$
with $h=x$ and $h=4x$.
