Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$ Finding the limit $\lim_{x\to\infty} (x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}$
I have tried using Taylor series. On a side not, I am not sure what to write when expanding both the $\sqrt{x^6+1}$ and the $e^{1/x}$. Do I write both $o(x^n)$ and $o(1/x^m)$? Seems like an abuse of notation.
Here is my attempt:
$(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}=$
$(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+\frac{1}{4!x^4}+o(\frac{1}{x^5}))-(1+\frac{x^6}{2}+o(x^6))$
But when I distribute terms there is no way for me to kill off the $x^n$ terms for positive $n$ and which makes me believe the limit is infinite. According to wolfram, the limit is $\frac{1}{6}$. 
What am I missing here? Is this notation acceptable?

Let $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6+1}=(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -\sqrt{x^6(1+\frac{1}{x^6})}$
This is $(x^3-x^2+\frac{x}{2})e^\frac{1}{x} -x^3\sqrt{1+\frac{1}{x^6}}$
$(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}+\frac{1}{4!x^4}+o(\frac{1}{x^5}))-x^3(1+\frac{1}{2x^6}+o(\frac{1}{x^{12}}))$
Expand your brackets and you're sorted. 
