Let f: (0,1] ->Rbe differentiable on (0,1], with |f(x)|<=1 for all x in (0,1]. For each n inN, let a(sub n)=f(1/n). Show that a(sub n), with n inN, converges.

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- Nov 11th 2008, 07:21 AMdhhnerdReal Analysis Problem
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. - Nov 12th 2008, 05:12 PMThePerfectHacker
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.