Let f: (0,1] -> R be differentiable on (0,1], with |f(x)|<=1 for all x in (0,1]. For each n in N, let a(sub n)=f(1/n). Show that a(sub n), with n in N, converges.
Let $\displaystyle f(x) = \sin \tfrac{1}{x}$ then $\displaystyle f$ is differenciable on $\displaystyle (0,\infty)$ and $\displaystyle |f|\leq 1$. The sequence is $\displaystyle a_n = f(\tfrac{1}{n}) = \sin (n)$.
However, $\displaystyle \{a_n\}$ is not a convergent sequence.