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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}

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  • $\begingroup$ I can't see how can you reduce the original limitto what you say, but since you're only interested in the asymptotics of the function then I think you can can do what you did in the last part, though I think it can be dangerous in some cases. $\endgroup$ – DonAntonio Mar 7 '16 at 15:06
  • $\begingroup$ In the first reduction I had the limit simplified to $\lim_{x\rightarrow\infty}[(x^2-3)/(x+2)]$ and due to the continuity of log, I brought the limit in ending with a $\infty/\infty$ situation thus applying L'Hopital since both equations are continuous and differentiable. $\endgroup$ – Andyvorld Mar 7 '16 at 15:13
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$$\lim_{x\to\infty}\log\frac{x^2-3}{x+2}=\lim_{x\to\infty}\left[\log\frac{x+\sqrt3}{x+2}+\log(x-\sqrt3)\right]=0+\infty=\infty$$

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