Let f(x)=x? Cos( 1/x), when x >; 0; F(x)=x, when x≤0. Discuss the derivability of f(x), and find f '(x) where it is derivable.

Solution: f (0) = 0; x→0？ limf(x)=x→0？ lim[x？ Cos( 1/x)]=0 because x→0? , cos( 1/x) is a bounded function, x? Is infinitesimal, so x→0? lim[x？ cos( 1/x)]= 0 = f(0)； ∴f(x) is continuous at point x=0.

And the left derivative at x=0 = f'(0? )= 1;

Right derivative at x=0 =f' (0 )=x→0？ limf '(x)=x→0？ lim[2xcos( 1/x)-x？ sin( 1/x)(- 1/x？ )]

=x→0？ Lim [2xcos (1/x)+sin (1/x)] does not exist;

Therefore, f(x) is continuous but nondifferentiable at x=0.