Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve:
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.

In the eternal words of Claude Leibovici, love Taylor Series.
We have 
\begin{align*}
\ln(x^{2}+2x+4) &= \ln(4) + \frac{x}{2} + \frac{x^{2}}{8} + \mathcal{O}(x^{3})\\
\ln(x+4) &=\ln(4)  + \frac{x}{4} -\frac{x^{2}}{32} + \mathcal{O}(x^{3}).
\end{align*}
So
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)} = \lim_{x\to 0} \frac{x}{\ln(4) + \frac{x}{2} + \frac{x^{2}}{8} - \left( \ln(4)  + \frac{x}{4} -\frac{x^{2}}{32}\right) + \mathcal{O}(x^{3})} = \lim_{x\to 0} \frac{x}{\frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$\lim_{x\to 0} \frac{x}{\frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})} = \lim_{x\to 0} \frac{\frac{1}{x} x}{\frac{1}{x}\left( \frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})\right)} = \lim_{x\to 0}\frac{1}{\frac{1}{4} + \frac{5x}{32} + \mathcal{O}(x^{2})} = \frac{1}{\frac{1}{4}} = 4.$$
