So, there is a way to prove it.Thanks, I'll try that!

[Poopsilon]

So I'd like to start out by asking why the interchanging of limits is phrased in terms of the property of uniform convergence? Can all double limits be written in terms of the limit of some sequence of functions and the limit as the input variable for these functions approaches some specific value?

For instance, consider this statement: Let $$f$$ be a continuous function defined on the interval $$[a,b]$$ such that $$f'$$ exists on $$(a,c)\cup(c,b)$$ and such that $$\lim_{x\rightarrow c} f'(x)$$ exists, that is to say when the left and right limits exist and are equal. Prove that $$f'(c)$$ exists.

Now I am fairly certain this is true and its a curiosity that came up in my homework while trying to stitch together a piece-wise function and prove it is everywhere infinitely differentiable.

The way I see it, what I need to show is this:

$$\lim_{x\rightarrow c} \left ( \lim_{t\rightarrow x} \frac{f(x) - f(t)}{x - t}\right ) = \lim_{t\rightarrow x} \left ( \lim_{x\rightarrow c} \frac{f(x) - f(t)}{x - t}\right )$$

So I need to exchange these two limits, neither of which is the limit of a sequence of functions. Any thoughts?

Edit: Actually now that I look at it I'm not entirely sure whether the right hand side of the equality is well-defined, since once the first limit is taken there will no longer exist an x to take t to, I can't decide whether I can consider x simply a dummy-variable in this case or not.. maybe it should just be:

$$\lim_{t\rightarrow c} \frac{f(c) - f(t)}{c - t}$$

Although that would require first proving that the derivative exists at c... which would prove my claim, although my claim seems to imply continuity of the derivative at that point as well.. hmm..

 

[micromass]

So I'd like to start out by asking why the interchanging of limits is phrased in terms of the property of uniform convergence? Can all double limits be written in terms of the limit of some sequence of functions and the limit as the input variable for these functions approaches some specific value?

For instance, consider this statement: Let $$f$$ be a continuous function defined on the interval $$[a,b]$$ such that $$f'$$ exists on $$(a,c)\cup(c,b)$$ and such that $$\lim_{x\rightarrow c} f'(x)$$ exists, that is to say when the left and right limits exist and are equal. Prove that $$f'(c)$$ exists.

Now I am fairly certain this is true and its a curiosity that came up in my homework while trying to stitch together a piece-wise function and prove it is everywhere infinitely differentiable.

The way I see it, what I need to show is this:

$$\lim_{x\rightarrow c} \left ( \lim_{t\rightarrow x} \frac{f(x) - f(t)}{x - t}\right ) = \lim_{t\rightarrow x} \left ( \lim_{x\rightarrow c} \frac{f(x) - f(t)}{x - t}\right )$$

So I need to exchange these two limits, neither of which is the limit of a sequence of functions. Any thoughts?

Edit: Actually now that I look at it I'm not entirely sure whether the right hand side of the equality is well-defined, since once the first limit is taken there will no longer exist an x to take t to, I can't decide whether I can consider x simply a dummy-variable in this case or not.. maybe it should just be:

$$\lim_{t\rightarrow c} \frac{f(c) - f(t)}{c - t}$$

Although that would require first proving that the derivative exists at c... which would prove my claim, although my claim seems to imply continuity of the derivative at that point as well.. hmm..

Indeed, it is not the exchanging of the limits that you need to prove, but it actually is

$$\lim_{x\rightarrow c} \left ( \lim_{t\rightarrow x} \frac{f(x) - f(t)}{x - t}\right ) = \lim_{x\rightarrow c} \frac{f(x) - f(c)}{x - c}\right )$$

I believe, one could do this with epsilon-delta definitions. But I think I have a nicer method:

Let

$$g(x)=\left\{\begin{array}{cc}f^\prime(x) & x\neq c\\ \lim_{x\rightarrow c}{f^\prime(x)} & x=c \end{array}\right.$$

Then g is a continuous function, that has a primitive function. This primitive function is continuous, and it can easily be shown that it will coincide with f, since f is also continuous...