Is $ f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} $ uniformly continuous?

Question

Is $$ f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} $$ uniformly continuous? I think it could be but can't prove it. Would appreciate the help. Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used. So, I'm left with the definition and Lipschitz

Answer 1

Let $x$ and $y$ be different points in $(1,+\infty)$. By the Mean Value Theorem, there exists a point $z$ between $x$ and $y$ such that

$$ f(x)-f(y)=\cos(\frac{1}{z})(\frac{1}{x}-\frac{1}{y})=\cos(\frac{1}{z})(\frac{y-x}{xy}) $$ Then $$ |f(x)-f(y)|\leq|x-y| $$ So $f$ is $1$-Lipschitz and, in particular, uniformly continuous.

Answer 2

$f^{\prime}(x)= (-1/ x^2)(\cos (1/x))$, which is bounded. So, given function is uniformly continuous.