How should I prove $\lim_{x \to \infty} \frac{1}{x^3} = 0$

   Use the definition of the limit to prove the following limit.

$$\lim_{x \to \infty} \frac{1}{x^3} = 0$$

This is my attempt at solving this question

Suppose $\epsilon > 0$, choose $M = \frac{1}{^3\sqrt{\epsilon}}$

Suppose $x>M$ $$\frac{1}{x}<\frac{1}{M}\ \text{(taking the reciprocal)}$$ $$\frac{1}{x^3}<\frac{1}{M^3}\ \text{(cubing both sides)}$$

Assuming $x > 0$ as the limit is as $x$ approaches $\infty$: $$\left|\frac{1}{x^3}\right|<\frac{1}{M^3}$$ $$\implies \left|\frac{1}{x^3} - 0\right|<\epsilon$$

I am unsure whether this is the right way to do so. I thought of this method after watching videos and reading up on limit proofs. I am self-learning all these topics purely for interest. Any corrections to my working will be greatly appreciated! Thank you.

Solution 1:

Well done, your logic works because giving a formula for $M$ based on $\epsilon$ which satisfies the proposition guarantees that such an $M$ exists for any $\epsilon,$ given that the formula is defined over the given domain. (which it is in this case)

This is a proof strategy called proof by construction and it's great for whenever you need to prove a given object exists. (as opposed to proof by contradiction, which is more useful for proving something is true for all objects of a certain type)

A few things worth noting, first off that cubing both sides of the inequality is justified because $f(x) = x^3$ is strictly increasing, which can be justified a number of ways but commonly by simply showing that $f'(x) = 3x^2$ is always positive. Second, the $x > 0$ can alternatively be justified with $M = \frac{1}{^3\sqrt{\epsilon}} > 0$ when $\epsilon > 0,$ so $x > M$ and $M > 0$ implies $x > 0.$