I was doing some Calculus homework the other day when I had a realization. As we know, the limit of sin(x) divided by x as x approaches 0 is 1:

$\underset { x\to 0 }{ lim } \frac { sin(x) }{ x } =1$

However, sometimes problems are presented to us that are similar but not exactly the same, such as this:

$\underset { x\to 0 }{ lim } \frac { sin(2x) }{ 3x } =??$

As I was doing Calculus problems, and I was repeatedly given situations like the above one, I began to notice a pattern:

$\underset { x\to 0 }{ lim } \frac { sin({ K }_{ 1 }x) }{ { K }_{ 2 }x } =\frac { { K }_{ 1 } }{ { K }_{ 2 } } $

I tested this numerous times with many different values, and this pattern always seemed to hold true. I then searched through my textbook (and online articles) to see if this was a standard identity that I had missed, but I couldn't find mention of it anywhere. Earlier today I asked my Calculus teacher about this pattern, and he also said that he'd never seen this before. It's hard for me to believe that this relationship/identity isn't already known (I'm just a high school student, after all. Some professor should have already found this!), so why isn't this taught more? Instead of being taught to memorize $\underset { x\to 0 }{ lim } \frac { sin(x) }{ x } =1$, why aren't we taught to memorize $\underset { x\to 0 }{ lim } \frac { sin({ K }_{ 1 }x) }{ { K }_{ 2 }x } =\frac { { K }_{ 1 } }{ { K }_{ 2 } } $? Is there some edge case or odd scenario where this identity doesn't work?


Is there a mathematical reason (for example, edge-cases where this doesn't work) why Calculus students, when they're learning about limits, are taught $\underset { x\to 0 }{ lim } \frac { sin(x) }{ x } =1$ instead of $\underset { x\to 0 }{ lim } \frac { sin({ K }_{ 1 }x) }{ { K }_{ 2 }x } =\frac { { K }_{ 1 } }{ { K }_{ 2 } } $?

Note: The answer to this question may be obvious, but I have done some research online and have not been able to find an answer. I'm just a high school student trying to figure out if there's a reason why this isn't taught (for example, is there an edge case where this doesn't work). Please let me know if more information is needed. Thanks in advance!

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    $\begingroup$ $\lim_{x\to 0}\sin(\alpha x)/\beta x = \alpha/\beta$ is an easy corollary of $\lim_{x\to 0}\sin x/x = 1$, you're supposed to be able to use laws of limits to figure out the former by your knowledge of the latter. $\endgroup$ – Stahl Oct 4 '16 at 21:51
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    $\begingroup$ In beginning math classes in high school, we are frequently taught information by fiat. Think about when you learned about logarithms. It requires some calculus to begin to properly define them, but in the meantime students gain valuable experience manipulating them and of course they are useful in chemistry and physics. The same is true for limits. Later in your calculus course you will learn how to do them properly, but for now you are told to memorize some results to allow you to gain experience solving various problems. In other words: you must be patient. $\endgroup$ – poweierstrass Oct 4 '16 at 21:53
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    $\begingroup$ Probably because 1) the simpler limit is the one needed for the proof of the formula of the derivative of $\sin x$, 2) the more general formula can be "easily" derived from the first and that we usually want to learn the simpler formulas even if we want to modify them subsequently. For example, we usually learn the formula for the derivative of $\sin x$ rather than that for $\sin f(x)$ because it's quite easy to apply the chain rule when needed. $\endgroup$ – Bernard Masse Oct 4 '16 at 21:55
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    $\begingroup$ @Rob First, you wrote the same expression in both limits, whereas Stahl wrote the correct case. Second, you can even make it simpler than substitution: $$\frac{\sin k_1x}x=k_1\frac{\sin k_1x}{k_1x}\xrightarrow[x\to0]{}k_1\cdot1=k_1$$ $\endgroup$ – DonAntonio Oct 4 '16 at 22:05
  • $\begingroup$ you could use L'Hopital's rule from real analysis - then clearly $\lim_{x\rightarrow 0}\frac{\sin K_1 x}{K_2 x}=\lim_{x\rightarrow 0}\frac{K_1\cos K_1 x}{K_2 }=\frac{K_1}{K_2}$ $\endgroup$ – MKF Oct 4 '16 at 22:05

I think it should be fairly easy to see that

$$ \lim_{x\to 0}\frac{\sin(K_1x)}{K_1x}=\lim_{y\to 0}\frac{\sin(y)}{y}=1 $$

setting $y=K_1 x$ and noting that $y\to 0$ as $x\to 0$. Then changing the bottom $K_1$ into a $K_2$ is nothing more than a constant multiplication by $K_1/K_2$, which gives your result.

As for why it's not taught, the prevailing taste in (many parts of) mathematics is to present results in the most economic fashion possible. Proving that $$\lim_{x\to 0}\frac{\sin(x)}{x}=1$$ requires some non-trivial arguments, whereas getting from there to $$\lim_{x\to 0}\frac{\sin(K_1x)}{K_2x}=\frac{K_1}{K_2}$$ is nothing more than some standard arguments using limits. So we present the 'hard' result, and leave the 'easy' corollary for the student to derive on the fly.

From a pedagogical point of view, the fact that you haven't been taught this result directly means that you've had some practice manipulating limits and have discovered it for yourself, which is far more valuable than being taught it by rote.

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    $\begingroup$ What you've said makes a lot of sense - thank you! It never occurred to me how easily you can derive $\underset{x\to0}{lim}\frac{sin({K}_{1}x)}{{ K}_{2}x}=\frac{{K}_{1}}{{K}_{2}}$ given $\underset{x\to0}{lim}\frac{sin(x)}{x}=1$ with just a few simple concepts (which both you and Doug M explained). This does raise some other questions to me, but they're not strictly math related, and this is not the place for those questions. Thank you for your explanation and help! $\endgroup$ – CodeIt Oct 5 '16 at 4:55
  • $\begingroup$ Of course the right-hand side of your first displayed equation should have $y \to 0$, not $x \to 0$. $\endgroup$ – LSpice Apr 4 '18 at 19:57
  • $\begingroup$ @LSpice Thank you $\endgroup$ – John Gowers Apr 5 '18 at 8:58

This is an important property of limits:

$\lim_\limits{x\to a} \frac {kf(x)}{g(x)} =k \lim_\limits{x\to a} \frac {f(x)}{ g(x)}$

This is another one:

$\lim_\limits{x\to 0} \frac {f(kx)}{g(kx)} =\lim_\limits{x\to 0} \frac {f(x)}{ g(x)}$

Your question is a specific example of the two of these more fundamental properties acting together:

$\lim_\limits{x\to 0} \frac {\sin(K_1x)}{K_2 x} = \frac {K_1}{K_2}\lim_\limits{x\to 0} \frac {\sin(K_1x)}{K_1 x} = \frac {K_1}{K_2}\lim_\limits{x\to 0} \frac {\sin(x)}{ x}$


Memorizing formulas vs Understanding the logic

Especially these days students are looking for formulas or clever mnemonics to solve problems. You do not need to memorize $$ \lim \limits_{x \rightarrow 0} \frac{\sin (K_1 x)}{K_2x} = \frac{K_1}{K_2} $$ when you know $$ \lim \limits_{x \rightarrow 0} \frac{\sin x}{x} = 1 $$

Instead of memorizing the first one, as Donkey 2009 explained, you can see it as a result of the second one.

Slightly related note

Recently, I saw an engineering student who uses a mnemonics to remember what $\sin 30^\circ$ is. I said you can easily derive it by drawing an equilateral triangle and one of the angle bisectors of it. He was surprised seeing this derivation. Instead of memorizing all $\sin 30, \sin 60$ etc, you can actually understand where they come from.

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