Evaluate this limit of integration in Complex analysis I am trying assignments of Complex analysis and I couldn't evaluate this particular integral.

Find the limit $\int_{|z|=R} |\sin(z)/z | |dz| $ as $R\to\infty$.

I am unable to see which result I should use and also there is confusion due to dz being in modulus and also $\sin(z) /z$.
Can you please tell significance of modulus here and how to compute the limit.
Thanks!!
First note that
$$
 \int_{|z|=R} \frac{|\sin(z)|}{|z|} |dz| = \int_0^{2 \pi} \frac{|\sin(R e^{it})|}{|R e^{it}|} R \, dt = \int_0^{2 \pi} |\sin(R e^{it})| \, dt \, .
$$
Generally, for $z= x+iy$,
$$
 |\sin(z)|^2 = \sin^2(x) + \sinh^2(y) \implies |\sin(z)| \ge |\sinh(y)| \, .
$$
If we restrict the integration to the arc $z = R e^{it}$ with $\pi/2 \le t \le 3\pi /2$ then
$$
|\sin(R e^{it})| \ge \sinh(R \sin(t)) \ge \sinh(\frac{R}{\sqrt 2}) \, .
$$
(This is what @reuns meant with the comment that “$|\sin(z)|$ is large near $iR$”).
It follows that
$$
\int_0^{2 \pi} |\sin(R e^{it})| \, dt \ge
\int_{\pi/2}^{3 \pi/2} |\sin(R e^{it})| \, dt \\
\ge \int_{\pi/2}^{3 \pi/2} \sinh(\frac{R}{\sqrt 2}) \, dt = \frac \pi 2 \sinh(\frac{R}{\sqrt 2}) \, ,
$$
and that grows to infinity for $R \to \infty$.
Hint : it is $\infty$.
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