Does differentiability imply convergence

Question

Can we say that if the limit of a sequence of functions is differentiable then the sequence is convergent?

I mean, I know that $\frac{\partial f(x,t)}{\partial x}$ exists. If I specify a sequence based on the definition of the derivative, something like $f_n(x,t)=f(x+1/n,t)$ do I know that this sequence converges just because the derivative exists? It seems so, but I am not sure.

If there's a limit for the sequence, it's convergent. – Apr 13 '14 at 21:13 — , Apr 13 '14 at 21:13
If the sequence $(x_n)$ converges to $x$, then, yes. – Apr 13 '14 at 21:13 — , Apr 13 '14 at 21:13

score 1 · Answer 1 · answered Apr 13 '14 at 21:42

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You just said the sequence has a limit, so yes, it's convergent.

answered Apr 13 '14 at 21:42

Jack M

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I edited my question. It will clarify what my doubt is – The Mighty Algernon Apr 13 '14 at 21:50

score 0 · Accepted Answer · answered Apr 13 '14 at 22:00

To answer the main question:

No, define $f_n(x,t)=(-1)^n$. We have $\frac{\partial f_n}{\partial x}=0$ for every $n$, and clearly this converges (uniformly) to a differentiable function $0$. However, it is rather obvious the sequence of functions is even pointwise convergent.

Does differentiability imply convergence

2 Answers2