<s>[INST] Find $\lim_{n\to\infty}\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$ $$\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+\frac{3}{n^3}\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+3\sqrt{\frac{4+n^6}{n^9}}}$$
$$=\frac{0+5}{2+0}$$
$$=\frac{5}{2}$$
...but the given answer is $1$. Where did I make a mistake??
 [/INST] In your answer, the term under the radical $\sqrt{4+n^6}$ does not play a real role anymore. Since this is potentially the "most complicated term", this situation is an hint (not a proof) that something has gone wrong with your proof. The exercise as lost most of its interest to test your skill.
If you  factor your expression correctly with the term in $n$ with the highest power, you get: 
$$\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}} = \frac{5n^3 \left(1+\frac{1}{5n}\right)}{2n^3+3n^3\sqrt{4/n^6+1}} = \frac{5n^3 \left(1+\frac{1}{5n}\right)}{n^3 (2+3\sqrt{4/n^6+1})}\,.$$
As for the notations, it is discouraged (even wrong) to write something like $\frac{1/n+5}{2+3/n} = \frac{5}{2}$, because it is not true. At the last stages, you should reintroduce the notion of limits, instead of writing a series of equalities.
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