$\lim_{x\rightarrow0^+}\frac{1}{x^n}$ (epsilon-delta proof)?

Question

Consider the following infinite limit

$\lim_{x\rightarrow0^+}\frac{1}{x^n}$

It suffices to show that the function has no upper bound given we approach the limit point from positive infinity, correct? Thus, can we write (?)

Given $\delta > 0$, then $x_0 < x < x_0 + \delta \implies f(x) > M$

\begin{align*} \frac{1}{x^n} &> M \\ \frac{1}{M^{\frac{1}{n}}} &> x \ \text{ if } \ \delta = \frac{1}{M^{\frac{1}{n}}} \end{align*}

Answer 1

You have the right idea, but to write a proper proof you should first fix a putative upper limit $M$ and show that there exists some $\delta$ such that $0 < x < \delta$ implies $x^{-n} > M$. In this case, a proof could go like this:

Let $M > 0$ and choose $\delta = M^{-1/n}$. Then $0 < x < \delta$ implies $ x^{-n} > \delta^{-n} = M$, as desired. Therefore, $\lim_{x \to 0^+} x^{-n} = \infty$.

In other words, you have correctly identified the steps to get $\delta$ from $M$, but when writing down the proof, you must go up your chain of equivalences instead of down.

Answer 2

Another attempt:

Note : $1/x^n >1/x$ for $1>x>0.$

Let $M$ be given.

Choose $\delta = \min(1,1/M)$.

$0 \lt x \lt \delta$ implies:

$ 1/x^n >1/x > 1/\delta \ge M.$