Evaluate $\lim_{x \to \infty} (e^{\frac{1}{x}}(x^3 - x^2 + \frac{x}{2}) - \sqrt{x^6 +1})$ using Taylor $\lim\limits_{x \to \infty}\left[e^{\frac{1}{x}}\left(x^3 - x^2 + \dfrac{x}{2}\right) - \sqrt{x^6 +1}\right]$
(I assume that we need to use $t = \frac{1}{x}$ in order to get lim, where argument goes to $0$.  But for now I'm not doing this since it won't help much at this moment).
$- \sqrt{x^6 + 1} = -\left( 1 + \dfrac{x^6}{2} + r_n (x)\right)$
Hence, we need somehow get $\frac{x^6}{2}$ from the $e^{\frac{1}{x}}(x^3 - x^2 + \frac{x}{2})$
Doesn't matter what I've tried, after expressing $e^{\frac{1}{x}}$ as Taylor polynomial max degree of any positive term is $3$.
Of course, I can multiply this term with $x^3$, but then I have infinity in nominator, which is not good. I don't know how to cancel $- \frac{x^6}{2}$ with something.
Also after multiplication of $e^{\frac{1}{x}}$ in polynomial form by $(x^3 - x^2 + \frac{x}{2})$ I get $x^3$ which, again, doesn't cancel with anything.
Multiply-divide by conjugate method helps a bit, but then there is "unsolvable" expression in denominator.
Am I doing something wrong?
If you've got another results, please, share

We have that
$$e^{\frac{1}{x}} = 1+\frac1x+\frac1{2x^2}+\frac1{6x^3} +o\left(\frac1{x^3}\right)$$
$$\sqrt{x^6 +1} = x^3\left(1+\frac1{x^6}\right)^\frac12= x^3+\frac12\frac1{x^3}+o\left(\frac1{x^3}\right)$$
and then
$$e^{\frac{1}{x}}\left(x^3 - x^2 + \frac{x}{2}\right) - \sqrt{x^6 +1}=$$
$$=x^3 - x^2 + \frac{x}{2}+x^2 - x + \frac{1}{2}+\frac x2 - \frac12 + \frac16- x^3+o(1)=$$
$$=\frac 1 6+o(1)\to \frac16$$
