<s>[INST] 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?
 [/INST] 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.
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