Determine by definition whether the function is Uniform continuity $f(x)=\dfrac{\sin(x)}{x}, x\in(0,\pi)$ Determine by definition whether the function is Uniform continuity
$$f(x)=\dfrac{\sin(x)}{x}, x\in(0,\pi)$$
$$|\dfrac{\sin(x)}{x}-\dfrac{\sin(y)}{y}|\leq|\dfrac{y\sin(x)-x\sin(y)}{xy}|\leq ??$$
I hardly succeed in how to solve this question thanks.
 A: Following the hint of the linked question we may write $$\left|\frac{\sin(x)}{x}-\frac{\sin(y)}{y}\right|=\left|\int_{0}^1\cos(xt)-\cos(yt) dt\right|\tag{1}$$
Using the identity
$$\cos(xt)=\cos\left(\frac{xt+yt}{2}\right)\cos\left(\frac{xt-yt}{2}\right)-\sin\left(\frac{xt+yt}{2}\right)\sin\left(\frac{xt-yt}{2}\right)$$ we may write eq. 1 as $$\left|\frac{\sin(x)}{x}-\frac{\sin(y)}{y}\right|=\left|2\int_{0}^1\sin\left(\frac{xt+yt}{2}\right)\sin\left(\frac{xt-yt}{2}\right) dt\right|\leq \left|2\int_0^1 dt\right|=2$$

Another way to justify the result is to first notice that $\left |\frac{\sin(x)}{x}\right |\leq 1$, from which it follows that $$\left |\frac{\sin(x)}{x}-\frac{\sin(y)}{y}\right |\leq \left|\frac{\sin(x)}{x}\right |+\left|\frac{\sin(y)}{y}\right |\leq 1+1=2$$
A: Using
$$ \cos A-\cos B=-2\sin\bigg(\frac{A+B}{2}\bigg)\sin\bigg(\frac{A-B}{2}\bigg),|\sin x|\le x$$
one has
\begin{eqnarray}
&&\left|\frac{\sin(x)}{x}-\frac{\sin(y)}{y}\right|\\
&=&\left|\int_{0}^1\cos(xt)-\cos(yt) dt\right|\\
&=&\left|-2\int_{0}^1\sin\left(\frac{xt+yt}{2}\right)\sin\left(\frac{xt-yt}{2}\right) dt\right|\\
&\leq& 2\int_{0}^1\left|\frac{xt-yt}{2}\right|dt\\
&\le&\frac12|x-y|
\end{eqnarray}
which implies that $\frac{\sin x}{x}$ is uniformly continuous.
