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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?

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  • $\begingroup$ Use l’Hospital’s rule. Result pops out from that. $\endgroup$
    – cdipaolo
    Sep 18, 2019 at 1:11
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    $\begingroup$ That's rather cyclical, don't you think? Taking the derivative requires the evaluation of this limit. $\endgroup$ Sep 18, 2019 at 1:42
  • $\begingroup$ It would make a great test question on a calculus 1 exam. $\endgroup$
    – Axion004
    Sep 18, 2019 at 2:38
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    $\begingroup$ @Axion004 that's what i'm in lol $\endgroup$
    – achandra03
    Sep 18, 2019 at 20:42

3 Answers 3

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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 $$\frac{n}{m} \cdot \lim_{x \to 0} \frac{\sin nx}{nx} = \frac{n}{m} \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = \frac{n}{m} \cdot 1 = \frac{n}{m}$$

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

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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

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

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