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.
 A: By L'Hospital we have 
$$\lim_{x\to 0} \frac{1}{\frac{2x + 2}{x^2 + 2x + 4} - \frac{1}{x+4}} = \color{red} 4$$
A: Outline: Flip it over, and calculate
$$\lim_{x\to 0} \frac{\ln(x^2+2x+4)-\ln(4)}{x} -\lim_{x\to 0}\frac{\ln(x+4)-\ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $\ln(x^2+2x+4)$ and $\ln(x+4)$ respectively. 
A: 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
$$\lim_{x\to 0} \frac{x}{\frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})} = \lim_{x\to 0} \frac{\frac{1}{x} x}{\frac{1}{x}\left( \frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})\right)} = \lim_{x\to 0}\frac{1}{\frac{1}{4} + \frac{5x}{32} + \mathcal{O}(x^{2})} = \frac{1}{\frac{1}{4}} = 4.$$
A: Rewrite it as
$$\lim\limits_{x\rightarrow0}\frac{x}{\ln \left( \frac{x^2+2x+4}{x+4}\right)}$$
Apply L'Hopital's rule:
$$\lim\limits_{x\rightarrow0}\frac{1}{\frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$\frac{1}{\frac{0^2+8\times0+4}{(0+4)(0^2+2\times0+4)}} = \frac{1}{\frac{4}{16}} = 4$$
A: An approach without L'Hopital's rule.
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}=\lim_{x\to 0} {1\over {1\over x}\ln {x^2+2x+4\over x+4}}=\lim_{x\to 0} {1\over \ln \big ({x^2+2x+4\over x+4}\big)^{1\over x}}$$
but 
$$({x^2+2x+4\over x+4}\big)^{1\over x}=({x^2+x+x+4\over x+4}\big)^{1\over x}=\big(1+{x^2+x\over x+4}\big)^{1\over  x}=\big(1+{ x(x+1)\over x+4}\big)^{1\over  x}=\big(1+\color\red x({x+1\over x+4})\big)^{1\over \color\red x}$$ which tends to $e^{1\over 4}$ as $x$ goes to $0$. So the original limit is ${1\over \ln{e^{1\over 4}}}=\color\red  4$
