Limit of $(\sin nx) / mx$ as $x$ approaches $0$ I was playing around with $$\lim_{x\to 0} \frac{\sin x}{x}$$ and experimenting with changing the coefficients on both the $x$'s. The limit of $(\sin 3x) / x$ is $3$ and the limit of $(\sin3x)/(2x)$ is $3/2$. Does this pattern hold true for all real numbers? That is, is the limit of $\cfrac{\sin nx}{mx}$ always $\cfrac{n}{m}$? If so, why?
 A: Observe that
$$\lim_{x \to 0} \frac{\sin nx}{mx} = \lim_{x \to 0} \left( \frac{n}{m}\cdot\frac{\sin nx}{nx} \right) = \frac{n}{m} \cdot \lim_{x \to 0} \frac{\sin nx}{nx}$$
Now, substitute $\theta = nx$ in the last limit. Finally, note that $\theta \to 0$ as $x \to 0$, and so
$$\frac{n}{m} \cdot \lim_{x \to 0} \frac{\sin nx}{nx} = \frac{n}{m} \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = \frac{n}{m} \cdot 1 = \frac{n}{m}$$
A: A bit more than the limit itself.
$$ y=\frac{\sin (nx)}{mx} =  \frac{n}{m}\cdot\frac{\sin (nx)}{nx}  = \frac{n}{m} \,  \frac{\sin (nx)}{nx}$$
Let $t=nx$ and use
$$\sin(t)=t-\frac{t^3}{6}+O\left(t^5\right) \implies \frac{\sin (t)}{t}=1-\frac{t^2}{6}+O\left(t^4\right)$$ Back to $x$
$$y=\frac{n}{m}\left(1-\frac {n^2}6 x^2\right)+O\left(x^4\right)$$
A: We have 
$$\lim_{x\to 0} \frac{\sin nx}{mx}$$
tends to the indeterminate form of $\frac{0}{0}$. We therefore need to apply L'Hopital's rule once to see that
$$\lim_{x\to 0} \frac{n\cos nx}{m}=\frac{n}{m}\lim_{x\to 0}\frac{\cos nx}{1}=\frac{n}{m}$$
