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?

$\text{HINT}$
$$x(16x^4 + x^2+1)^{1/4}-2x^2 $$ $$=x^2(16+ \frac{1}{x^2}+\frac{1}{x^4})^{1/4}-2x^2$$ $$=x^2(\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2)$$ $$=\frac{\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2}{\frac{1}{x^2}}$$
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 $\frac{1}{32}$
