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.
 A: 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
$$
\frac1{4}\left| \frac{x-\tfrac{1}{2}}{x^3+\tfrac{1}{2}} \right|\leqslant \frac1{4}\left| \frac{x}{x^3} \right|\leqslant \frac1{4M^2}\leqslant \epsilon 
$$
∎
A: One thing I'll say before I fully answer is that, when students are first learning proofs, many will often have a tendency to insert symbols wherever and whenever they can. I did the same thing myself many years back, thinking that by using symbols everywhere, I was being more precise and mathematical, but remember that at the end of the day, a human being has to read and verify your proof, and it's much easier to read a proof written in plain old english (or whatever language your reader knows).  The definition of a limit that is relevant in your case is the following: We say $\lim_{x \to \infty} f(x) = L$ if for every $\epsilon > 0$, there exists a real number $N > 0$ such that if $x > N$, then $|f(x) - L| < \epsilon$.
Let's fix $\epsilon > 0$. We first note that because $\lim_{x \to \infty} \frac{1}{x^2} = 0$, there is a $N > 0$ such that if $x > N$, then $\lvert \frac{1}{x^2} \rvert < \epsilon$. If you haven't yet proven that the previous limit is true using the definition of a limit, you should do so now before moving on with this example. Let's now observe the following: for $x > \max(N,1)$, the following will be true:
$$\left| \frac{x^3 + x}{4x^3 + 2} - \frac{1}{4}\right| = \left| \frac{2x-1}{8x^3 + 4}\right| < \left| \frac{2x-1}{8x^3}\right| < \left| \frac{2x}{8x^3} \right| = \left| \frac{1}{4} \frac{1}{x^2}\right| = \frac{1}{4}\left|\frac{1}{x^2} \right| < (1/4)\epsilon < \epsilon$$
where the first inequality follows by the fact that $8x^3 + 4$ is greater as a denominator than $8x^3$.
A: First note that $$\frac{x^3+x}{2+4x^3}-\frac{1}{4} = \frac{4(x^3+x)-(2+4x^3)}{4(2+4x^3)} = \frac{4x-2}{16x^3+8} = \frac{\frac{1}{4x^2}-\frac{1}{8x^3}}{x^3+\frac{1}{2}}$$
and so $$\left|\frac{x^3+x}{2+4x^3}-\frac{1}{4}\right| \leq \frac{\left|\frac{1}{4x^2}\right|+\left|\frac{1}{8x^3}\right|}{\left|x^3+\frac{1}{2}\right|} = \frac{\frac{1}{4x^2}+\frac{1}{8x^3}}{x^3+\frac{1}{2}}.\tag{1}\label{eq1}$$
Now let $\epsilon>0$. Using that $$\lim_{x\to\infty}\frac{1}{x^n} = 0$$ for any integer $n$, we know that there exists $r_1$ such that $x> r_1$ implies that $$\frac{1}{4x^2}<\frac{1}{x^2}<\epsilon.$$
Similarily you can find  an $r_2$ such that $x>r_2$ implies that $$\frac{1}{8x^3}<\frac{1}{x^3}<\epsilon.$$
Note that we can pick $r_1$ and $r_2$ so that both are greater than or equal to one. This is handy, because then e.g. if $x>r_1$ one has that the denominator $x^3+1/2$ is also greater than $1$.
Now let $r=\max\{r_1,r_2\}$. What can you conclude about the right hand side of \eqref{eq1} if $x>r$?
