sin(3x) over 5x

Idea. Adjust the expression so that the argument of the sine and the denominator are the same.

Computation. For $x\neq0$,
$$\frac{\sin(3x)}{5x}=\frac35\cdot\frac{\sin(3x)}{3x}.$$
Set $u=3x$. When $x\to0$, $u\to0$, so $\dfrac{\sin u}{u}\to1$. Hence the limit is $\dfrac35\cdot1=\dfrac35$.

Remark. In general $\dfrac{\sin(ax)}{bx}\to\dfrac ab$ as $x\to0$.