It should use Laurent's series, but the limit converges to $\infty$, hence there should be the substitute, but I don't know what to put as $t$:
$$\lim_{x \to +\infty}\Big[\Big(x^3-x^2+ \frac{x}{2}\Big)e^\frac{1}{x}-\sqrt{x^6+1}\Big]$$
How to solve the limit
0
$\begingroup$
calculus
limits
laurent-series
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0Well... try $t=\frac1x$. – 2017-02-23
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0what do you mean with $$\infty+$$? – 2017-02-23
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0Actually $+\infty$, fixed a typo – 2017-02-23
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0If you expand $e^{\frac{1}{x}}$ as $1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+o(\frac{1}{x^3})$ and $\sqrt{x^6+1}$ as $x^3+o(1)$, then you should be able to obtain the correct limit, $\frac{1}{6}$. – 2017-02-23