Let $f:(0,1) \rightarrow \mathbb R$ be continous. Suppose that $|f(x) - f(y)| \leq |\sin x - \sin y |$ for all $x,y\in (0,1)$. Then 
*

*$f$ is discontinuous at least one point in $(0,1)$

*$f$ is continuous everywhere on $(0,1)$, but not uniformly continuous on $(0,1)$

*$f$ is uniformly continuous on $(0,1)$

*$\lim_{x \rightarrow 0^+} f(x)$ exists.
Since  $\sin x$ is uniformly continuous on $(0,1)$, because its derivative is bounded., So $f$ is uniformly continuous on $(0,1)$. So (1) and (2) are false, (3) is true, but what can i say about (4).
Please check first three option and give me hint about (4).
Thank you 
 A: Since $\sin' = \cos$, we see that $|\sin'(x)| \le 1$ and so
$|\sin x - \sin y | \le |x -y|$.
Hence $|f(x) -f(y)| \le |x-y|$.
It follows that 1,2 are false and 3,4 are true.
To elaborate 4:

 Note that if $x_n$ is Cauchy, then so is $f(x_n)$. Furthermore, if $x_n,x'_n$ are Cauchy with the same limit, then it is straightforward to show that $f(x_n), f(x'_n)$ have the same limit. Now consider Cauchy $x_n$ that converge to zero.

A: Since $f(x)$ is uniformly continuous on $(0,1)$, by continuous extension theorem, it is defined at its end points i.e $0$ and $1$ respectively. Since $f$ is continuous, the limit exists.
A: (1)-(3) look fine. (Instead of using the fact that $\sin x$ has bounded derivative, you could also use the fact that it's continuous on the compact interval $[0, 1]$ and thus uniformly continuous there, and hence uniformly continuous on the subspace $(0, 1)$.) As for (4), note that $f$ is bounded, so there exists a sequence $x_n \to 0$ such that $f(x_n)$ converges to some limit $\alpha.$ The estimate
$$|f(x) - \alpha| \leq |f(x) - f(x_n)| + |f(x_n) - \alpha| \leq |\sin x - \sin x_n| + |f(x_n) - \alpha|$$
then shows that $\lim_{x\to 0^+} f(x) = \alpha$.
