9
$\begingroup$

Let $f:[a,+\infty) \to \Bbb{R}$ be a differentiable function with the property: $$\inf\{f'(x)|x>a\}>0$$Prove that $\lim_{x \to +\infty}f(x)=+\infty$

Here is my solution:

Let $c=\inf\{f'(x)|x>a\}>0$ and $x \geq a+1$

From Mean Value Theorem exists $x_0 \in [a+1,x]$ such that $$f(x)=f(a+1)+f'(x_0)(x-a+1)$$

But $f'(x) \geq c,\forall x \geq a+1$

Thus $$f(x) \geq f(a+1)+c(x-a+1),\forall x \geq a+1 \Rightarrow \liminf_{x \to +\infty}f(x)=+\infty$$

Thus $\lim_{x \to +\infty}f(x)=+\infty$

Is my solution correct or am i missing something?

Thank you in advance.

$\endgroup$

2 Answers 2

7
$\begingroup$

I think your argument is right. From a writing perspective, instead of using $a+1$ I would choose $b>a$ and the inequalities would look a little bit nicer; but I guess that's a matter of taste.

$\endgroup$
0
$\begingroup$

There are still a few steps to get to the conclusion that the limit is infinity, although they are trivial enough that your instructor will likely let them slide. You need to show that given any M, there exists x0 such that if x>x0 then f(x) > M. So you can take x0 = (M-f(a))/c + a. You don't need the MVT, you can just use the Fundamental Theorem of Calculus:

f(x) = [f(x)-f(a)]+f(a)

The portion in the brackets is equal to the integral from a to x of f'(x), so

f(x) =

[integral(f'(x)) from a to x] +f(a) >

c*(x-a)+f(a)

Since x > x0 = (M-f(a))/c + a and c>0, we can substitute that in for x above and preserve the inequality:

f(x) >

c*(M-f(a))/c + a - a)+f(a) =

c*(M-f(a))/c +f(a) =

M-f(a)+f(a) =

M

So f(x) > M, QED.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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