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Consider a sequence of functions $f_n$ where $f_n : \mathbb{R} \to \mathbb{R}$ and $f_n$ are all differentiable with derivatives $f^\prime_n$. The sequence $f_n$ and the sequence $f^\prime_n$ both converge uniformly to functions $f$ and $g$ respectively. According to the definition given in this wiki page on uniform convergence in section 'To Differentiability"

If $ f_n $ converges uniformly to $ f $, and if all the $ f_n $ are differentiable, and if the derivatives $f^\prime_n$ converge uniformly to $g$, then $ f $ is differentiable and its derivative is $g$.

Q.1) Can a similar definition be used for higher order derivatives ?

This is a different definition of derivative which is not same as the usual definition f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.

Q.2) In case of both not being in agreement with each other which one should I use for further analysis on $f$ for example in theorems involving the derivative of $f$ ?

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    $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ is the definition. Part of the conclusion of the theorem is that $f'$ exists, and that $f'=g$. That is, $f'$ is not defined to be $g$; it already has a definition, when it exists, and the theorem asserts that under those hypothesis is does exist, and it also equals $g$.2011-01-21

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http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability

If $f_n$ converges uniformly to $f$, and if all the $f_n$ are differentiable, and if the derivatives f'_n converge uniformly to $g$, then $f$ is differentiable and its derivative is $g$.

So if you use the theorem and the derivates f'_n are also differentiable you have again that the f'_n converge uniformly to f' and you can recursively use the theorem from above (you will need that f''_n converges uniformly to some $g'$).

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The theorem you mention certainly applies to higher order derivatives, exactly as written by user3123 above.

For your other question, it seems to me that you are worried about the situation where $f$ may not be differentiable, but there exists a sequence $f_n$ which uniformly converges to $f$, with each $f_n$ differentiable and the derivatives f_n' converging uniformly to some function $g$.

Do you have an example of such a situation?

To rephrase what Jonas mentioned already in his comments, this is not likely to happen. When the sequence $f_n$ uniformly converges to $f$, and the derivatives f_n' uniformly converge to $g$, the following holds:

$\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h} = g(x).$

This is the derivative of $f$. It is not an alternative definition, it really is a theorem. I'm sorry! Perhaps you can give some more details on the exact situation you have in mind.

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    @Qiachu Yuan : I found my mistake...thank you...I guess converse need'nt be true...it is not a necessary condition for $f$ to be differentiable.2011-01-22