# Infinite limit as X tends to infinity

## Homework Statement

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

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
Homework Helper
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)

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?

Last edited:
tiny-tim
Homework Helper
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!

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 :)

Last edited:
tiny-tim
… So I noticed that the bigger x gets, the closer $$cos\frac{1}{x}$$ gets to 1.