# Why does $\lim\limits_{\theta\to\infty}n\sin\frac {\theta}n=\theta$

Questions:

1. Why do we have$$\lim\limits_{\theta\to\infty}n\sin\dfrac {\theta}n=\theta\tag1$$
2. How do we prove $(1)$?

I started off with the well known limit:$$\lim\limits_{\theta\to 0}\dfrac {\sin\theta}{\theta}=1\tag2$$ And substituted $\theta:=\dfrac \theta n$ into $(1)$ to get$$\lim\limits_{\frac \theta n\to 0}\dfrac {\sin\dfrac \theta n}{\dfrac \theta n}=\lim\limits_{\frac \theta n\to0}\dfrac {n\sin\frac \theta n}{\theta}\tag3$$ However after that, I'm not sure what to do. Apparently, the RHS of $(1)$ has a limit of $1$, so that the limit of $n\sin\frac \theta n=\theta$.

• Do you mean $\lim_{n\to\infty}$? Commented Feb 15, 2017 at 1:14
• It is false that $\lim_{\theta\to\infty}\dots=\theta$, since $\theta$ itself goes to infinity, but $\sin$ is bounded. Perhaps you meant the limit as $n\to\infty$? Commented Feb 15, 2017 at 1:15
• The expression $\lim_{\theta\to\dots}\dots = \theta$ isn't even a valid equation because $\theta$ isn't a free variable of the left hand side. Commented Feb 15, 2017 at 1:26

I think you meant $\lim_{n\to \infty}n \sin\left(\frac{\theta}{n}\right) = \theta$, which you basically proved is true:

$\lim_{n\to \infty}n \sin\left(\frac{\theta}{n}\right) = \lim_{n\to \infty} \left( \theta\cdot \frac{\sin\left(\frac{\theta}{n}\right)}{\frac{\theta}{n}} \right) = \theta \cdot \lim_{n\to \infty} \left( \frac{\sin\left(\frac{\theta}{n}\right)}{\frac{\theta}{n}} \right) = \theta \cdot 1 =\theta$

Assuming you meant $\lim_{n\to\infty}$, note that by letting $u=\theta/n$, we get

$$\lim_{n\to\infty}n\sin\frac\theta n=\theta\lim_{u\to0^{\pm}}\frac{\sin u}u=\theta\times1$$

where the side we approach depends on the sign of theta, but it doesn't affect the limit.

Or, with equivalents, $\sin\dfrac\theta n\sim_\infty \dfrac\theta n$, so $$n\sin\frac\theta n\sim_\infty n\frac\theta n=\theta.$$

The original expression does not seem correct to me, let's just substitute $$n$$ for $$\theta$$ and rewrite $$\lim_{n\to \infty}n \sin\frac{\theta}n$$ At this point you can easily deduce that $$\lim_{n\to \infty}\Biggl(\frac{\sin\frac{\theta}n}{\frac1n}\Biggl)\space=\space\lim_{n\to \infty}\Biggl(\frac{\sin \frac{\theta}{n}}{\frac{\theta}{n}}\theta\Biggl)$$ Since we know that argument of sinus function will be ridiculously close to zero, we can use below expression to make things easier $$\sin \frac{\theta}{n}\space\approx\space\frac{\theta}{n}$$ If we make required substitutions, we'll eventually get $$\lim_{n\to \infty}\frac{\frac{\theta}{n}}{\frac{\theta}{n}}\theta\space=\space\theta$$