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Can anyone give a hint? $$\lim_{x\to+\infty}\left(\left(x^3+\frac{x}2-\tan \frac1{x}\right)\text{e}^{\frac1{x}}-\sqrt{1+x^6}\right)$$

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What are your (failed?) attempts at finding the limit of this problem? – Clayton Feb 22 '13 at 4:32
Once again, a question from you with absolutely no context or no thoughts of your own. Despite the fact that you've been asked numerous times to improve your questions. (This is a quote from a comment on another of your questions.) – Did Feb 27 '13 at 7:37

More general hint: expand $\tan\frac{1}{x}$ and $e^{1/x}$ in series at infinity.

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$e^{1/x} \to 1+1/x+O(1/x^2)$, $\sqrt{1+x^6} = x^3\sqrt{1+x^{-6}} \to x^3(1+x^{-6}/2 + O(x^{-12})) = x^3 + x^{-3}/2 + O(x^{-9}) $, and $\tan(1/x) \to 1/x + O(1/x^2)$.

Combining these, $\begin{align} (x^3+x/2-\tan(1/x))e^{1/x}-\sqrt{1+x^6} &\to (x^3+x/2-1/x+O(1/x^2))(1+1/x+O(1/x^2)) \\ &-(x^3 + x^{-3}/2 + O(x^{-9}))\\ &=x^3+x/2-1/x+O(1/x^2) + x^2+1/2-1/x^2+O(1/x^3)+O(x) \\ &-(x^3 + x^{-3}/2 + O(x^{-9}))\\ &= x^2+x/2+1/2 + O(x) \\ &= x^2+O(x) \\ \end{align} $

Essentially, the $x^3$ cancel out, and what is left is primarily the $x^2$ plus smaller terms (unless I made a mistake). The $x/2$ and $\tan(1/x)$ don't seem to matter.

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