Using the definition of a limit, prove that lim
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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