Using the definition of a limit, prove that $$\lim_{n \rightarrow \infty} \frac{2}{n^4} = 0$$
My take: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$
$$\left |\frac{2}{n^4} - 0 \right | < \epsilon$$
Notice that $\frac{2}{n^4} < \frac{2}{n} < \epsilon$, so I think I can say $n > 2\epsilon$ (not sure if I can do this).
So, given all that rough work,
Proof: Given $\epsilon >0$, choose $N > 2\epsilon$ and suppose $n \ge N$ such that
$$\left | \frac{2}{n^4} \right | < \frac{2}{n} \le \frac{2}{N} \le \frac{1}{(2\epsilon)^2} = \epsilon?$$
I'm sure I screwed it up somewhere. Can someone help me please?
Answer
I continue on your proof.
Your take: I want to prove that given $\epsilon > 0$, there $\exists N \in \mathbb{N}$ such that $\forall n \ge N$
$$\left |\frac{2}{n^4} - 0 \right | < \epsilon$$
Let $\epsilon>0$, we want to find an $N$ such that $\forall n\in\mathbb{N},~ n\ge N\Longrightarrow\left |\frac{2}{n^4}\right | < \epsilon$.
Observe on this inequality: $\left |\frac{2}{n^4}\right | < \epsilon$. Since the "$n$" we discuss is only under natural numbers, so we roughly assume that $n\in\mathbb{N}$. Then we can go on our observations:
$$\begin{alignat}{2}
&\left |\frac{2}{n^4}\right | < \epsilon\\
\underset{~n\in\mathbb{N}}{\Longleftrightarrow}&\frac{2}{n^4}<\epsilon\\
\Longleftarrow &~\frac{2}{n^4}<\underbrace{\frac{2}{n}<\epsilon}_{\text{hope it can be true}}
\end{alignat}
$$
Hence, from the "$\Longleftarrow$", when $\frac{2}{n}<\epsilon$, then $\left |\frac{2}{n^4}\right | < \epsilon$ can be true. But when will $\frac{2}{n}<\epsilon$?
$$\begin{alignat}{2}
&\frac{2}{n}<\epsilon\\
\Longleftrightarrow&2
\end{alignat}
$$
So until now, we have gain that $\displaystyle\forall n\in\mathbb{N},~n>\frac{2}{\epsilon}\Longrightarrow\left|\frac{2}{n^4} - 0 \right| < \epsilon$.
Then we choose $N$ to be $\lceil\frac{2}{\epsilon}\rceil$, then $\forall n\geq N,~\left|\frac{2}{n^4} - 0 \right| < \epsilon$. $\square$
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