Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?

The limit is not infinity. If you need to do it the hard way, consider
$$
x(\sqrt[4]{16x^4+x^2+1}-2x)
=x\frac{\sqrt{16x^4+x^2+1}-4x^2}{\sqrt[4]{16x^4+x^2+1}+2x}
=\frac{x(x^2+1)}{(\sqrt[4]{16x^4+x^2+1}+2x)(\sqrt{16x^4+x^2+1}+4x^2)}
=\frac{x^3(1+\frac{1}{x^2})}
 {x^3
  \Bigl(\sqrt[4]{16+\frac{1}{x^2}+\frac{1}{x^4}}+2\Bigr)
  \Bigl(\sqrt{16+\frac{1}{x^2}+\frac{1}{x^4}}+4\Bigr)
 }
$$
However, it's much easier with a Taylor expansion. If you substitute $x=1/t$, with easy algebraic manipulations the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt[4]{16+t^2+t^4}-2}{t^2}=
2\lim_{t\to0^+}\frac{\sqrt[4]{1+t^2/16+t^4/16}-1}{t^2}
$$
and
$$
\sqrt[4]{1+t^2/16+t^4/16}=1+\frac{1}{4}\frac{t^2}{16}+o(t^2)
$$

About the case when the limit is infinity, it depends on the conventions you're following whether the limit “exists”. Check with your instructor.
