# 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?

• Use l’Hospital’s rule. Result pops out from that. Sep 18, 2019 at 1:11
• That's rather cyclical, don't you think? Taking the derivative requires the evaluation of this limit. Sep 18, 2019 at 1:42
• It would make a great test question on a calculus 1 exam. Sep 18, 2019 at 2:38
• @Axion004 that's what i'm in lol Sep 18, 2019 at 20:42

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 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 $$x$$ $$y=\frac{n}{m}\left(1-\frac {n^2}6 x^2\right)+O\left(x^4\right)$$

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

$$\lim_{x\to 0} \frac{n\cos nx}{m}=\frac{n}{m}\lim_{x\to 0}\frac{\cos nx}{1}=\frac{n}{m}$$