derivative of limit function diverges from limit of derivatives

For a function sequence, one cannot always change the of http://planetmath.org/node/6209taking limit and differentiating (http://planetmath.org/Differentiate), i.e. it may well be

$$\underset{n\to \mathrm{\infty }}{lim}\frac{d}{dx}{f}_n(x)\ne \frac{d}{dx}\underset{n\to \mathrm{\infty }}{lim}{f}_n(x),$$

in the case that a sequence of continuous (and differentiable) functions converges uniformly; cf. Theorem 2 of the parent entry (http://planetmath.org/LimitFunctionOfSequence).

Example.  The function sequence

$f_n(x):={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{x^3}{{(1+x^2)}^j}}=x-{\displaystyle \frac{x}{{(1+x^2)}^n}}\mathit{\hspace{1em}\hspace{1em}}(n=\mathrm{\hspace{0.33em}1},\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots })$(1)

provides an instance; we consider it on the interval  $[-1,\mathrm{\hspace{0.17em}1}]$.  It’s a question of partial sum the converging geometric series

$$\frac{x^3}{1+x^2}+\frac{x^3}{{(1+x^2)}^2}+\frac{x^3}{{(1+x^2)}^2}+\mathrm{\dots }$$

(although one cannot use Weierstrass’ criterion of uniform convergence).  Sincethe limit function is

$$f(x):=\underset{n\to \mathrm{\infty }}{lim}\left(x-\frac{x}{{(1+x^2)}^n}\right)=x\mathit{\hspace{1em}}\forall x\in [-1,\mathrm{\hspace{0.17em}1}],$$

we have

$$\underset{[-1,\mathrm{\hspace{0.17em}1}]}{sup}|f_n(x)-f(x)|=\underset{[-1,\mathrm{\hspace{0.17em}1}]}{sup}\frac{|x|}{{(1+x^2)}^n}⟶0\mathit{\hspace{1em}}\text{as}n\to \mathrm{\infty },$$

which means by Theorem 1 of the parent entry (http://planetmath.org/LimitFunction) that the sequence (1) converges uniformly on the interval to the identity function.  Further, the members of the sequence are continuous and differentiable.  Furthermore,

$$f_n^{\prime }(x)=\mathrm{\hspace{0.33em}1}-\frac{1+(1-2n)x^2}{{(1+x^2)}^{n+1}},$$

whence

$$\underset{n\to \mathrm{\infty }}{lim}{f}_n^{\prime }(x)=\mathrm{\hspace{0.33em}1}\mathit{\hspace{1em}}(x\ne \mathrm{\hspace{0.33em}0}).$$

But in the point  $x=0$  we have

$$\underset{n\to \mathrm{\infty }}{lim}{f}_n^{\prime }(0)=\underset{n\to \mathrm{\infty }}{lim}0=\mathrm{\hspace{0.33em}0},$$

which says that the limit of derivative sequence of (1) is discontinuous in the origin.  Because

$$f^{\prime }(x)\equiv \mathrm{\hspace{0.33em}1},$$

we may write

$$\underset{n\to \mathrm{\infty }}{lim}\frac{d}{dx}{f}_n\ne \frac{d}{dx}\underset{n\to \mathrm{\infty }}{lim}{f}_n.$$