# Help me prove this: If fn is differentiable on X subset of R for every n and fn(x)  goes to f(x) on X, then f is differentiable on X.*the fn is equivalent to f sub n XD (i don't know how to type...

Help me prove this: If fn is differentiable on X subset of R for every n and fn(x)  goes to f(x) on X, then f is differentiable on X.

*the fn is equivalent to f sub n XD (i don't know how to type it here XD)

* this is for my Real Analysis class

degeneratecircle | Certified Educator

Are you sure you weren't asked to prove it or find a counterexample, because this isn't true as stated. Consider the functions

`f_n(x)=x^n` defined on [0,1]. Each `f_n` is differentiable everywhere and `f_n->f` where

`f(x)={(0 if 0<=x<1),(1 if x=1):}`

This isn't continuous at `x=1` and so isn't differentiable there, providing a counterexample. However, it is true that if `{f_n}` is a sequence of functions differentiable on `[a,b]` and such that `{f_n(x_0)}` converges for some `x_0 in [a,b]`, and also if `{f'_n}` converges uniformly on `[a,b]` , then `{f_n}` converges to a differentiable function.

That proof would take up too much space here, but it can be found in chapter 7 of Rudin's Principles of Mathematical Analysis (3rd ed) if you're interested.

mayomigem | Student
thank you for pointing that out :D we have to give a counterexample for that :D