How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$ with the definition of $\lim\limits_{x \to \infty}f(x)$? How do I show $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$  with the definition of $\lim\limits_{x \to \infty}f(x)$?
What I have done so far:
$\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} =\lim\limits_{x\to\infty}{\frac{x^{3}(1+\tfrac{1}{x^2})}{x^3(2\tfrac{1}{x^3}+4)}}=\frac{1+0}{0+4}=\frac{1}{4}$
At this point I wanted to use the definition of $\lim\limits_{x \to \infty}f(x)$
Given: $f : D \subset \mathbb{R} \rightarrow \mathbb{R}$
$\exists r \in \mathbb{R}$ with $(r,\infty) \subset D$.
for $\eta \in \mathbb{R}$ define:
$\lim\limits_{x\to\infty}{f(x)}=\eta : \Leftrightarrow  \forall \epsilon >0\ \exists \delta > r \forall x > \delta : f(x) \in B_{\epsilon}(\eta) $\
to show: $\lim\limits_{x\to\infty}{\frac{x^{3}+x}{2+4x^3}} = \frac{1}{4}$
let $\epsilon >0$
$\exists \delta>6  \ \forall x > \delta: \left | \frac{x^{3}+x}{2+4x^3}-\frac{1}{4} \right |<\epsilon$
The plan was to go on and find $\delta$ depending on $\epsilon$ but I didn't get it.

You have that
$$
\left | \frac{x^{3}+x}{2+4x^3}-\frac{1}{4} \right |=\left | \frac{x^{3}+x}{4(\tfrac{1}{2}+x^3)}-\frac{1}{4} \right |=\frac1{4}\left| \frac{x-\tfrac{1}{2}}{x^3+\tfrac{1}{2}} \right|
$$
Then, if you define $M:=\max\{1,\tfrac{1}{2\sqrt{\epsilon }}\}$ then for $x\geqslant M$ we have that
$$
\frac1{4}\left| \frac{x-\tfrac{1}{2}}{x^3+\tfrac{1}{2}} \right|\leqslant \frac1{4}\left| \frac{x}{x^3} \right|\leqslant \frac1{4M^2}\leqslant \epsilon 
$$
∎
