Passing the limit through the derivative of a differentiable function

[kbgregory]

Homework Statement

Suppose f is differentiable on an open interval I and let x* \in I. Show that there exists a sequence {x_n}\subset I such that lim[n->inf]{x_n}=x* and lim[n->inf]{f'(x_n)}=f'(x*).

Homework Equations

We know that a function g is continuous iff for any sequence {x_n} with lim[n->inf]{x_n}=x*, lim[n->inf]{g(x_n)}=g(x*).

The Attempt at a Solution

I think I need to show that since f is differentiable on I, then its derivative is continuous on I, and since its derivative is continuous on I, then there exists a sequence {x_n} with lim[n->inf]{x_n}=x* for which lim[n->inf]{f'(x_n)}=f'(x*).

But I am not sure how to show this, or even if its right.

 

[snipez90]

No, f differentiable on I does not imply its derivative is continuous. The canonical counterexample is f(x) = (x^2)*sin(1/x) if x =/= 0 and f(x) = 0 if x = 0.

In the conclusion, you have a double limit. There is a certain procedure to apply to these limits that would make the problem very simple. The continuity of f would be essential in this case.

But I think it's probably more illuminating to work from first principles. Write out the definition of the derivative of f at x* (epsilon-delta definition of limit), pick a specific sequence converging to x*, and make the appropriate modifications to your definition to show the conclusion.

 

[GVAR717]

I did what snipez90 said, but I got stuck. I know that it is necessary to apply the Mean Value Theorem, but I don't really understand how. Maybe you can figure out how the MVT can be applied.

 

[JG89]

Take your x \in I. You know that there is a sequence of positive numbers, h_n such that \lim_{n \rightarrow \infty} h_n = 0. Thus \lim_{n \rightarrow \infty} \frac{f(x + h_n) - f(x)}{h_n} = f'(x).

Now look at the difference quotient [f(x + h_n) - f(x)] / h_n . By the MVT [f(x + h_n) - f(x)] / h_n = f'(c_n) for x < c_n < x + h_n

As n goes to infinity that difference quotient gets closer and closer to the derivative of f at x, right? So what can you say about the sequence f'(c_n)?