Infinite limit as X tends to infinity

[IdanH14]

Homework Statement

I am required to express in \varepsilon - \delta way what I'm suppose to prove in case lim_\below{(x \rightarrow \infty)} f(x) = \infty

Homework Equations

None.

The Attempt at a Solution

So first, intuitively I thought that what this means is that f(x) is bigger than any arbitrary number when x is bigger than any arbitrary number. So I attempted to combine the \varepsilon - \delta definitions of when x tends to infinity and when limit f(x) tends to infinity.

I came up with this:
lim_\below{(x \rightarrow \infty)} f(x) = \infty if for every M>0 there exists N>0 so that for every x>M, f(x)>N.

I am unsure of whether it's the correct definition. Anyone can verify that?

 

[tiny-tim]

Welcome to PF!

Hi IdanH14! Welcome to PF!

(have a delta: δ and an epsilon: ε and an infinity: ∞ )

lim_\below{(x \rightarrow \infty)} f(x) = \infty if for every M>0 there exists N>0 so that for every x>M, f(x)>N.

Yes , except it's the other way round …

no matter how large N is, we can find an M above which f(x) > N.

(you get the same result if you use 1/f, 1/δ, and 1/ε)

 

[IdanH14]

Thanks! :)
Let me summarize it to see if I got it. It should be
For every N>0 there exists M>0, so that for every x>M, f(x)>N
Right?

 

[tiny-tim]

Thanks! :)
Let me summarize it to see if I got it. It should be
For every N>0 there exists M>0, so that for every x>M, f(x)>N
Right?

Right!

 

[IdanH14]

I like you usage of icons. I think I'll adopt it. ;)

Now, I'm trying to solve an exercise in my math book with this principle. Despite the fact I already learned infinite limits arithmetics, I'm required to prove that lim_\below{x\rightarrow \infty}) x cos\frac{1}{x} = \infty in this cumbersome way. So I think I have a solution, but again, my insecurities creep in.

So I noticed that the bigger x gets, the closer cos\frac{1}{x} gets to 1. If I'll choose M=N then, unless they N=\infty I'll get something that's smaller than N, because X is multiplied by something which is close to 1, but doesn't equal 1. But if M=2N then problem eliminated. Almost.

Why almost? Because if \frac{1}{x}>1 then cos\frac{1}{x} is potentially getting farther from 1. So, what I was thinking was to say M=max(1,2N) and then problem solved.

I hope I'm clear enough. Is this a valid proof? Did I find for every N>0 an M>0 that meet the requirements?

Thanks :)

 

[tiny-tim]

… So I noticed that the bigger x gets, the closer cos\frac{1}{x} gets to 1.

That's correct.

I can work out why you're then using M = 2N (because of course M = N doesn't quite work), but you haven't actually specified why M = 2N does work.

(for example, does it work for any N?)