I was wondering about the steps to show that the following limit does not exists. $$\lim_{x\rightarrow\infty}[\log(x^2-3)-\log(x+2)]$$ I know that by using L'Hopital's Rule and the continuity of logarithms I can reduce the limit to: $$\lim_{x\rightarrow\infty}[\log(2x)] \text{ which does not exists}$$
But since that $$x^2-3 \sim x^2,\text{as } x \rightarrow \infty $$ $$x+2 \sim x,\text{as } x \rightarrow \infty $$ would it be possible to show that the original limit does not exists by replacing the two asymptotically equivalent function into the logs (due to the continuity of logarithms). To result with: \begin{align} \lim_{x\rightarrow\infty}[\log(x^2-3)-\log(x+2)] & = \lim_{x\rightarrow\infty}[\log(x^2)-\log(x)]\\ & = \lim_{x\rightarrow\infty} [2\log(x)-\log(x)]\\ & = \lim_{x\rightarrow\infty} [\log(x)]\\ \end{align}
$$\lim_{x\to\infty}\log\frac{x^2-3}{x+2}=\lim_{x\to\infty}\left[\log\frac{x+\sqrt3}{x+2}+\log(x-\sqrt3)\right]=0+\infty=\infty$$