How to compute $\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$ through Taylor series I am trying to compute the below limit through Taylor series:
$\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$
What I have already tried is first of all change the variable x to
$x=1/t$ and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change $e$ to exponent polynomial up to t=6 + $O(X^6)$
however, I don`t know what can I do with square root.  

HINT:
$$\sqrt{x^6+3}=\frac1{t^3}\left(1+3t^6\right)^{1/2}=\frac1{t^3}+\frac{\frac12\cdot3t^6}{t^3}-\frac{\frac1{2\cdot4}\cdot(3t^6)^2}{t^3}+\frac{\frac{1\cdot3}{2\cdot4\cdot6}\cdot(3t^6)^3}{t^3}-\cdots\\(2x^3-2x^2+x)e^{1/x}=\left(\frac2{t^3}-\frac2{t^2}+\frac1t\right)\left(1+t+\frac{t^2}{2\cdot1}+\frac{t^3}{3\cdot2\cdot1}+\cdots\right)$$
