Find $$ \lim_{x\to-\infty}\ln\left(\sqrt{x^2+4}+x\right). $$
I got this limit which gives me $\ln(0^+)=-\infty$. Is this ok? I ended up with my answer in this way: my limit is equal to $$\ln\left(\lim_{x\to-\infty}\left(\frac{4}{\sqrt{x^2+4}-x}\right)\right)=\ln0^+=-\infty$$