How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$

Question

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule?
I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}\frac{\cos(ah)a}{1} = a$$ but I don't know how to behave without that way

Answer 1

Hint:

$$\frac{\sin(ha)}{h} = a\cdot\frac{\sin(ha)}{ha}$$

Also, remember what $$\lim_{x\to 0}\frac{\sin(x)}{x}$$ is equal to?

Answer 2

Notice that $\sin(x) = \sum_{k = 0}^\infty (-1)^k \frac{x^{2k + 1}}{(2k + 1)!}$ for all $x \in \mathbb R$. Thus, $$ \frac{\sin(ah)}{h} = \frac 1 h \sum_{k = 0}^\infty (-1)^k \frac{(ah)^{2k + 1}}{(2k + 1)!} = \sum_{k = 0}^\infty (-1)^k \frac{a^{2k + 1} h^{2k}}{(2k + 1)!} \longrightarrow a$$ as $h \to 0$ by dominated convergence for example.