Given a function $f:\mathbb R \rightarrow \mathbb R $ such that $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$, prove $f(x)\sin(x)$ is not uniformly continuous.

I tried to go by definition:

  • $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$ $\ \Rightarrow$ $\ \forall M>0 \ \exists E>0\ \forall x>E:f(x)>M$
  • Proving the function is not uniformly continuous: $\ \exists \epsilon>0\ \forall\delta>0\ \exists x_1,x_2\ \in\mathbb R \ s.t. |x_1-x_2|<\delta:|f(x_1)-f(x_2)|\ \geq\epsilon$
  • $|f(x_1)\sin(x_1)-f(x_2)\sin(x_2)|> ?$

How can I continue from here?


Put $\epsilon = 1$.

Put $x_1 = 2 \pi n$, $x_2 = x_1 + \delta/2$.

$|f(x_1) \sin(x_1) - f(x_2) \sin(x_2)| = |f(x_2) \sin(\delta/2)|$.

For sufficiently large n, this quantity is greater than $\epsilon$

  • 1
    $\begingroup$ Concise and complete (+1) $\endgroup$ – Mark Viola Feb 17 '17 at 20:49

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