1. ## Limits

I have got a problem like this :

Let f : R -> R be a function such that there exists constants b, m > 0 and c belongs to R such that
f(x) > mx + c for all x > b

using only the definition of the limit we have to prove that limit of f(x) when x goes to inifinity = inifinity

I have tried to prove it like this :

For all N > 0 if we can find some 'a' > 0 such that x > a implies (mx + c) > N
then that in turn implies f(x) > N

So that we can say limit of f(x) when x goes to inifinity = inifinity

But the problem is I can't find an 'a' > 0 for this.

I have tried to write it in reverse to find an 'a'

i.e mx + c > N
x > (N-c)/m

But here I get stuck coz 'c' can be either >= N or <N

What I need to do is a rigorous proof.

2. ## Re: Limits

consider a sequence Xn of numbers > b that tends to infinity.then (mXn+c) also tends to infinity and since f(Xn) is > mXn +c it is obvious that it also tends to infinity...

3. ## Re: Limits

Just start with: Given $N>c$, ...

Since $N$ can be arbitrarily large, you may not set an upper bound, but you are allowed to set a lower bound.

4. ## Re: Limits

I have another problem regarding this.. This comes as the 2nd part of the above problem.

Let f be differentiable on (0,infinity) using the mean value theorem and the result proven in the above question prove that of f and f' both are strictly increasing on (0,infinity) then limit of f(x) when x goes to infinity equals infinity.

Well as f is differentiable on (0,infinity) we can say f is continuous on (0,infinity).. therefore we can apply the mean value theorem for this.

If M is an arbitrarily large real number we can say

there exist some 'c' in (0,M) s.t f'(c) = [f(M)-f(0)]/M > 0 (as f is strictly increasing)

From there on I can't find a way to rigorously prove this problem :-(

So any help would be so great.. :-(

5. ## Re: Limits

Originally Posted by Kristen111111111111111111
Well as f is differentiable on (0,infinity) we can say f is continuous on (0,infinity).. therefore we can apply the mean value theorem for this.
You are correct that since $f$ is differentiable on $(0,\infty)$, it is continuous on the same interval. However, what is required to apply the Mean Value Theorem (I will abbreviate MVT from now on)?

The statement of the MVT requires that $f$ be continuous on a closed and bounded interval and differentiable on the interior of the same interval (the same interval without its endpoints is an open interval). So, you cannot apply the MVT to (0,M). Instead, you need a closed interval. Well, let's create a closed interval. Choose arbitrary points $0. We know $f(x)$ is continuous on $[x_0,x_1]$ and differentiable on $(x_0,x_1)$, so we can apply the MVT on this interval.

Originally Posted by Kristen111111111111111111
there exist some 'c' in (0,M) s.t f'(c) = [f(M)-f(0)]/M > 0
So, after you apply the MVT on this interval, you get some $b$ as you stated (only I am using different interval to find it and a different variable). To state a little more formally, by the MVT, there exists $b \in (x_0,x_1)$ such that $f'(b) = \dfrac{f(x_1)-f(x_0)}{x_1-x_0}$. Since we chose $x_0$ and $x_1$ with $x_0 and we are given that $f(x)$ is strictly increasing, this implies that both the numerator and denominator are positive. Hence, their quotient is positive. So, we now have a point $b$ with $f'(b)>0$. Let $m = f'(b)$ and $c = f(b)-bf'(b)$. You have $m>0, b>0$, so if you can show that $f(x)>mx+c$ for all $x>b$, then you can apply the first result to prove this 2nd part of your problem.