<s>[INST] 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. 
 [/INST] Hint:
You can evaluate it based on this limit, $$\lim_{x\to0}\frac{\ln(x+1)}x=1,$$ by writing,
$$\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}=\lim\limits_{x \to 0} \frac{\ln(1+x)}{x}\cdot\frac{4x+x^2}{\ln(1+4x+x^2)}\cdot\frac{x}{4x+x^2}=\ldots
$$
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