How to calculate the limit of $\frac{\sin(ax)}{x}$ for $x\to0$

you can rewrite $\dfrac{\sin (ax)}{x} = a \dfrac{\sin (ax)}{ax}$, then take limits, as suggested by @rscwieb in the comments

This is easy if you know the power series of the sine function.

$$\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} +\cdots$$

Then $$\sin(ax) = a x - \frac{a^3 x^3}{6} + \frac{a^5 x^5}{120} -\cdots$$

and $\frac{\sin(ax)}{x} = a - \frac{a^3 }{6}x^2 + \frac{a^5 }{120}x^4 -\cdots$ for all $x \neq 0$. Since power series are continuous you can find the limit for $x \rightarrow 0$ by simply setting $x=0$ in this expression, so the limit is $a$.