Editing Uniform convergence (section)

===To differentiability===
If <math>S</math> is an interval and all the functions <math>f_n</math> are [[derivative|differentiable]] and converge to a limit <math>f</math>, it is often desirable to determine the derivative function <math>f'</math> by taking the limit of the sequence <math>f'_n</math>. This is however in general not possible: even if the convergence is uniform, the limit function need not be differentiable (not even if the sequence consists of everywhere-[[analytic function|analytic]] functions, see [[Weierstrass function]]), and even if it is differentiable, the derivative of the limit function need not be equal to the limit of the derivatives. Consider for instance <math>f_n(x) = n^{-1/2}{\sin(nx)}</math> with uniform limit <math>f_n\rightrightarrows f\equiv 0</math>. Clearly, <math>f'</math> is also identically zero.  However, the derivatives of the sequence of functions are given by <math>f'_n(x)=n^{1/2}\cos nx,</math> and the sequence <math>f'_n</math> does not converge to <math>f',</math> or even to any function at all.  In order to ensure a connection between the limit of a sequence of differentiable functions and the limit of the sequence of derivatives, the uniform convergence of the sequence of derivatives plus the convergence of the sequence of functions at at least one point is required:<ref>Rudin, Walter (1976). ''[[iarchive:PrinciplesOfMathematicalAnalysis|Principles of Mathematical Analysis]]'' 3rd edition, Theorem 7.17.  McGraw-Hill: New York.</ref>

: ''If <math>(f_n)</math> is a sequence of differentiable functions on <math>[a,b]</math> such that <math>\lim_{n\to\infty} f_n(x_0)</math> exists (and is finite) for some <math>x_0\in[a,b]</math> and the sequence <math>(f'_n)</math> converges uniformly on <math>[a,b]</math>, then <math>f_n</math> converges uniformly to a function <math>f</math> on <math>[a,b]</math>, and <math> f'(x) = \lim_{n\to \infty} f'_n(x)</math> for <math>x \in [a, b]</math>.''