Show that if is a continuous real-valued function on the interval and , then
Let . It should be fairly clear that . Thus (if it exists) is less than or equal to m.
To get the reverse inequality, let . The continuity of f ensures that there is some interval of length on which . Then . Take the n'th root and let to see that . Finally, take the limit as .
I do not want to argue against Opalg because I am a noob compared to him in analysis , but I offer a suggestion.
I think this problem is better solved using the concepts of limsup's and liminf's.
Let be the length of the interval. Let .
Opalg hath shown us that .
Then he says by .
But we need to know the limit exists.
So a way around this problem is to perhaps say first take and then .
But the problem is depends on and so how we know that as ?
Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because and so . However, . Thus, we have shown but . This means that is a convergent sequence with .
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That is quite correct. The efficient way to explain the solution is to use limsup in connection with the upper estimate, and liminf for the lower estimate.
I didn't do that, because I didn't want to deter anyone who might be unfamiliar with limsup and liminf. So I just included a parenthetical comment to the effect that you need to establish that the limit exists.