Limit $\lim_{x\to∞}{(x^3-x^2+{x \over2})e^{1\over x}-\sqrt{x^6-1}}$

Question

Could someone please help me with this limit, I know i should use Taylors series but I only mess up the limit and don't get nowhere .... I spent an hour with this problem, with no result :(

$\lim_{x\to∞}{(x^3-x^2+{x \over2})e^{1\over x}-\sqrt{x^6-1}}$

Answer 1

Note that

$$e^{1\over x}=1+\frac1x+\frac1{2x^2}+\frac1{6x^3}+o\left(\frac1{x^3}\right)$$

$$\sqrt{x^6-1}=x^3\left(1-\frac1{x^6}\right)^\frac12=x^3\left(1-\frac1{2x^6}+o\left(\frac1{x^6}\right)\right)=x^3-\frac1{2x^3}+o\left(\frac1{x^3}\right)$$ thus

$${(x^3-x^2+{x \over2})e^{1\over x}-\sqrt{x^6-1}}=x^3+x^2+\frac{x}{2}+\frac1{6}-x^2-x-\frac12+\frac{x}2+\frac12-x^3+o\left(1\right)=\frac16+o(1)\to\frac16$$

Answer 2

Write the limit as $\lim_{t\to 0}t^{-3}\Big((1- t+\frac{t^2}{2})e^t-\sqrt{1-t^6}\Big)$ and then use Taylor's expansion of order $3$ at zero for the stuff inside the big bracket. There, you will have some cancellations.

In other words, use $e^t = 1+t+\frac{t^2}{2}+\frac{t^3}{6}+o(t^3)$ and $\sqrt{1-t^6} = 1-\frac{1}{2}t^6+o(t^8).$