Questions:
- Why do we havelim
- 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.
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