If the derivative is bigger than positive constant, is the limit at infinity infinity

Hello

I'm trying to prove the following: if is differentiable on all and there exists a real positive constant such that > c for all , then the limit of f at infinity is infinity.

I've managed to prove that if the derivative is positive, then by the Mean Value Theorem the function must be strictly increasing. But I can't seem to manage to prove that the limit isn't some real number...

try this: can you prove that if f'(x) > c then

http://quicklatex.com/cache3/ql_df0a...f1affe2_l3.png


The RHS clearly does not converge to an upper limit

By the Mean Value Theorem, for every x > 0 we have

( f( x ) - f ( 0 ) ) / x = f ' ( t_x ) >= c

That is,

f ( x ) >= c x + f ( 0 )

Now, take limits as x -> + infty .


Edited: Sorry, I didn't see the previous post.

Thanks to both of you...but I'm not a hundred percent sure why the RHS doesn't converge. Is it because k is a variable and can be as big as we want? If L was a limit of f then f + c would just converge to L + c, right?

Also...might it be enough to say that if f'(x) > c, then f(x) > c*x + k ?

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Thanks to both of you...but I'm not a hundred percent sure why the RHS doesn't converge. Is it because k is a variable and can be as big as we want?

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yes

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If L was a limit of f then f + c would just converge to L + c, right?

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Your notation is causing the confusion. F(x) + c converges to L+c, but f(x+c) converges to the same limit as f(x) (i think!).

http://quicklatex.com/cache3/ql_e9f1...dff02e2_l3.png

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Also...might it be enough to say that if f'(x) > c, then f(x) > c*x + k ?

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i dont know how rigorous you're expected to be. It can be shown robustly using the mean value theorum (as demonstrated by Fernando for the case where k=0). of course, it would be easier to use fernandos result directly as he suggested in his post.



Fernondos solution is better, but i thought this was an interesting alternative approach to the problem:

Spoiler:

if f'(x)>c then we can say f'(x)=c + u(x) where u(x) >0

Now consider f(x1) and f(x2), for arbitrary x1 and x2:

http://quicklatex.com/cache3/ql_9788...a2213ca_l3.png

The integral on the RHS is positive (no need to actually evaluate it!) since u(x) > 0

Taking the limit as x2 tends to infinity gives the required result for f(x2), and hence f(x). i think.

Okay, that looks clever, but I'm afraid I haven't learned much about integrals yet so I'm not quite equipped to understand it...I'll go over it again once I know more :) Thanks anyway