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.

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:
$$\frac{1}{\frac{0^2+8\times0+4}{(0+4)(0^2+2\times0+4)}} = \frac{1}{\frac{4}{16}} = 4$$
