<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] Observe that 
$$g(x) = 16x^4 + x^2+1 = 16(x^2 + \frac 1{32})^2 + \frac{63}{64}$$
Take the fourth root to get 
$$\sqrt[4]{g} \sim 2 \sqrt{x^2 + \frac 1{32}}$$
as $x \to \infty$.
We still have to multiply by $x$ and then subtract $2x^2$. We see that 
$$x \cdot 2 \sqrt{x^2 + \frac 1{32}} = 2 \sqrt{x^4 + \frac {x^2}{32}} = 2 \sqrt{(x^2 + \frac 1{64})^2 -\frac{1}{4096}}$$
And $$2 \sqrt{(x^2 + \frac 1{64})^2 -\frac{1}{4096}} \sim 2(x^2 + \frac 1{64}) = 2x^2 + \frac 1{32}$$ as $x \to \infty$.
Finally we subtract the $2x^2$ to get $$2x^2 + \frac 1{32}- 2x^2 = \frac 1{32}$$
Hence  $$\lim_{x\to \infty} x \sqrt[4]{16x^4 + x^2+1}-2x^2 = \frac 1{32}$$ which can be verified using L'Hôpital's Rule.
But your follow-up question was important, so I'll answer it as if the limit were infinity.
When we write $$\lim_{x\to \infty}f(x)=\infty$$ we are saying that for any real number $M$, there is a real number $\delta$ (which will depend on $M$) such that $f(x)\gt M$  for all $x$ such that $0\lt |x| \lt \delta$. It is important to realize, however, that the limit does not exist. For the limit to exist, it needs to be a number, and $\infty$ is not a number. The expression "$\lim_{x\to \infty}f(x)=\infty$" is just a concise way to express the idea that $f$ increases without bound.
</s>