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?
 A: $\text{HINT}$
$$x(16x^4 + x^2+1)^{1/4}-2x^2 $$ $$=x^2(16+ \frac{1}{x^2}+\frac{1}{x^4})^{1/4}-2x^2$$ $$=x^2(\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2)$$ $$=\frac{\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2}{\frac{1}{x^2}}$$
You can multiply the last fraction with $$\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}+2$$ both numerator and denominator and then you will multiply the resulting fraction with $$\sqrt{16 + \frac{1}{x^2}+\frac{1}{x^4}}+4$$ both numerator and denominator.
For more convinience with your calculations put $u=\frac{1}{x^2} \to 0$ as $x \to +\infty$
So calculate the limit of the function $f(u)$ as $u \to 0$ where $u=\frac{1}{x^2}$ 
Thus you will get the result $\frac{1}{32}$
A: 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.
A: 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.
A: Let's make use of the (not so) popular formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ Putting $x^{2}=1/t$ we can see that $t\to 0^{+}$ and the expression under limit is transformed into $$2\cdot\frac{(1+t/16+t^2/16)^{1/4}-1}{t}=2\cdot\frac{(1+t/16+t^2/16)^{1/4}-1}{t/16+t^{2}/16}\cdot\left(\frac{1}{16}+\frac{t}{16}\right)$$ and therefore the limit is equal to the limit of $$\frac{1}{8}\cdot\frac{(1+t/16+t^2/16)^{1/4}-1}{t/16+t^{2}/16}$$ as $t\to 0^{+}$. Next we put $u=1+t/16+t^2/16$ so that $u\to 1^{+}$ and the desired limit is equal to $$\frac{1}{8}\lim_{u\to 1^{+}}\frac{u^{1/4}-1^{1/4}}{u-1}$$ which equals $(1/8)(1/4)=1/32$ via formula $(1)$.
