# Prove that there are no maximum and minimum of $\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$

The problem I'm in stuck is

For $$\forall x_1, x_2\in\mathbb R$$, prove that $$\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$$ has neither maximum nor minimum.

I tried to fix $$x_2$$ to make a single variable function $$f(x):=\frac{\sin(x)-\sin(x_2)}{x-x_2}$$ and find the point $$x_0$$ such that $$f'(x_0)=0$$. I wanted to find some properties of $$f(x_0)$$, so that if $$x_2$$ varies, I wanted to show that $$g(x_2):=f(x_0(x_2))$$ has no maximum and minimum. One that I found is $$\frac{\sin(x_0)-\sin(x_2)}{x_0-x_2}=\cos(x_0)$$ so it somehow implies that $$f(x_0)=\cos(x_0)$$, which is nothing helpful to me. How can I make the direction from here? Thanks in advance.

$$\frac {\sin x_1-\sin x_2} {x_1-x_2} \leq 1$$ whenveer $$x _1 \neq x_2$$ by MVT. Also, taking $$x_2=0$$ and letting $$x_1 \to 0$$ we see that the expresion takes value as close to $$1$$ as we want. Hence, the supremum is $$1$$. To show that this supremum is never attained note that if $$x_1>x_2$$ and $$\frac {\sin x_1-\sin x_2} {x_1-x_2} = 1$$ implies that $$1=\frac 1{ x_1-x_2}\int_{x_2}^{x_1} \cos t \, dt\leq \frac 1{ x_1-x_2}\int_{x_2}^{x_1} dt=1$$. This forces equality to hold through oot so $$\cos t=1$$ for all $$t \in (x_2,x_1)$$. This is obviously a contradiction. So the maximum value is not attained . A similar argument works for the minimum.