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.

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 ${1\over \ln{e^{1\over 4}}}=\color\red  4$
