Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble figuring how to rigorously prove limits as $x \to \infty$. For instance, how would I rigorously prove that

$$\lim_{x \to \infty} x^2 = \infty$$

I am stumped. It is, of course, obvious. However, I do not think simply stating such a fact qualifies are a proper proof. Please note that a link to a resource that explains this would suffice as help.

share|improve this question
    
Pick some number $N.$ Is $x^2>N$ for sufficiently large $x$? –  Andrew Sep 25 '12 at 23:58

2 Answers 2

up vote 6 down vote accepted

Let's play a game. The rules are as follows:

  • I'm going to pick a number $N$.
  • You're going to pick a number $M$.
  • I'm going to pick a number $x$ bigger than $M$.

If $x^2 > N$, you win. Otherwise, I win.

Do you think you'll win this game? If so, tell me what your strategy is.

share|improve this answer
    
This is the answer I wanted to write, but didn't. –  MJD Sep 26 '12 at 2:14
    
I wonder if it is ever ok to have $M$ (or $\delta$ depending on how you name it) depend on $x$. For example, "pick $\delta = 2x$". Of course that is not necessary in this case, but would it ever be? –  providence Sep 26 '12 at 3:16
    
@providence No, that would be against the rules, like White demanding to see Black's reply to his opening move before making it. But your question is very insightful, because in other cases one can vary the order and it makes an important difference. For example, a function $f$ is said to be continuous if, whenever I choose $x$ and $\epsilon$, and then you choose $\delta$, and I choose $y$ with $|y-x|<\delta$, then you win if $|f(y)-f(x)|<\epsilon$. But $f$ has the much stronger property of being uniformly continuous if you win even if I may postpone my choice of $x$ to the very end. –  MJD Sep 26 '12 at 4:27

You say "Let $N>0$ be given. I will now show that whenever $x$ is big enough, $x^2$ is bigger than $N$. To do this, I will find an $\delta>0$ such that whenever $x>\delta$, I will be able to show that $x^2>N$."

The interesting part is the details of how you produce the required $\delta$, which will vary from problem to problem. In this case one way to do it is to take $\delta = N$, because then whenever $x>\delta$, you know that $x^2 > \delta^2 = N^2 > N$. (Unless $N\le 1$, in which case taking $\delta = 1$ will do fine.)

share|improve this answer
    
...well, except when $N \leq 1$. :-) –  cardinal Sep 26 '12 at 1:57
    
Thanks; I have corrected the post. –  MJD Sep 26 '12 at 2:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.