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Dad here, helping his kid out with math.

I get it that to determine the derivative of $\sin(x)$ you do:

$$\frac{\sin(x+h) - \sin(x)}h = \frac{\cos(x)\sin(h)}h \to \cos(x)$$

However I assume that $\sin(h) = h$ for $h\to 0$. This correct because the "angle" (not sure what the correct word is in English) of $\sin(x) $ is $1$ for $x=0$. But this is what I'm trying to prove in the first place. So a circular reference. How to solve this?

Calvin Khor
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ADBF
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    Hi! Welcome to MSE! Just to confirm, you want to show that $\frac{\sin x}{x} ~ 1 $ for very small angles $x$? – P. J. Mar 26 '21 at 11:05
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    Thanks for taking the time. I think so. In order to determine the derivative of sin(x), I use the derivative of sin(x). I do this when I assume that sin(h) = h, so indeed that sin(h)/h = 1 – ADBF Mar 26 '21 at 11:09
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    Yes, I get your point. Using Taylor series or L'Hopital will be circular, as the order of logic goes as : limits >> continuity >> derivatives. Does this post help you? – P. J. Mar 26 '21 at 11:14
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    Hi, I typeset the mathematics in your question and made the title descriptive (I took a minor "leap of faith" when I added an arrow $\to$ to the centered equation, I hope it is what you meant.) If you want to do it yourself in future posts, you can get started here, and a more complete reference can be found here. – Calvin Khor Mar 26 '21 at 11:15
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    Thank you both for your kind help. The post mentioned gives plenty to contemplate, I will need some time to digest it all, PJ. I will use the typesetting as well in the future, thanks for pointing it out,Calvin. – ADBF Mar 26 '21 at 11:23
  • A little comment to make here, but $\sin 0 = 0$ , whereas $\cos 0 = 1$. –  Mar 26 '21 at 11:23
  • You may try this for the kid: Draw a circle of radius $r$. Connect centre $O$ to a point $P$ on the circle using straight line OP. Now take the arc PH that subtends an angle of $h$ at the centre. Note that when $h$ is small enough, arc PH and straight line PH are indistinguishable and that PH is perp. to OP. Lenght of arc PH =radius times angle at centre= rh= length of straight line PH, when $h$ is small enough. What is $\sin h$ now? Perpendicular/hypoteneous $=\sin h=rh/r=h$. Do you see it now why? – Koro Mar 26 '21 at 11:31
  • You are missing a step: $\sin(x+h)-\sin(x)=\sin x(\cos h -1) + \cos x \sin h$, so you also have to show that $(\cos h - 1)/h\to 0$ as $h\to 0$. Easier to use @ZAhmed's method. – TonyK Mar 26 '21 at 15:13

2 Answers2

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The fact that $\sin(h) \approx h$ for small $h$ is usually proven non-circularly (although circles are involved :D) by resorting to the following inequalities $$\begin{split} \sin (h) &\leqslant h \leqslant \tan (h),\\ \cos (h) &\geqslant \frac 1 2, \end{split} \tag 1$$ both valid when $0 < h < \pi/3$. You can see they hold by drawing a unit circle; here's a helpful picture for the first one:

$\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}$enter image description here

Indeed, when $0 < h < \pi/3$, we obtain, by rearranging terms in inequality $(1\text a)$ that $$0 \leqslant \sin (h) - h \leqslant \tan (h) - \sin (h ). \tag2$$ However, by the definition of $\tan$ and the trigonometric identity $1-\cos(h) = 2\sin^2(h/2)$, $$\tan (h) - \sin (h )= \frac{\sin (h)}{\cos (h)}(1-\cos (h)) = \frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right). $$ By employing inequality $(1\text a)$ again in the forms $\sin(h/2) \leqslant h/2$ and $\sin(h) \leqslant h$, and inequality $(1\text b)$ for the cosine at the bottom, we see that $$\frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right) \leqslant \frac h {1/2} \cdot 2 \left(\frac h 2\right)^2 = h^3. \tag 3$$ Merging inequalities $(2)$ and $(3)$ entails that $0\leqslant \sin(h) - h \leqslant h^3$, that is (adding $h$ to all sides) $h \leqslant \sin(h) \leqslant h^3 + h$, that is (dividing everything by $h$) $$1 \leqslant \frac{\sin(h)}{h} \leqslant 1+ h^2,$$ which ultimately implies that $$\lim_{h\to 0^+} \frac{\sin(h)}{h} = 1 $$ by the squeeze theorem (both $1$ and $1+h^2$ clearly tend to $1$ as $h\to 0^+$).

The complete limit (as $h\to 0$ from both sides) follows from the fact that $\sin(-h) = -\sin (h)$.

giobrach
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  • I know I'm supposed to accept one answer, but I'd be bluffing if would do that. My math skills are to limited to appreciate your answer fully, especially when the Big O comes in. I'm sorry about that, but I thank you for trying to answer me. – ADBF Mar 26 '21 at 16:40
  • Hello! If you have specific doubts, I am available for any question concerning any part of my answer. If necessary, I will edit it to include more information or rephrase critical bits. – giobrach Mar 26 '21 at 16:42
  • $\theta$ is the angle subtended at the centre or $\theta$ is the arc length? Please describe my why so? – Math-Learner Mar 26 '21 at 17:31
  • It is both: the circle has radius $1$ – giobrach Mar 26 '21 at 17:32
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    @ADBF: I had mentioned similar thing in my comment:) – Koro Mar 27 '21 at 05:23
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It should rather be $$\lim_{h \to 0}\frac{\sin(x+h) - \sin(x)}h =\lim_{h \to 0} \frac{2 \cos (x+h/2) \sin(h/2)}{h} =\cos x \lim_{h \to 0} \frac{\sin(h/2)}{h/2}=\lim_{h \to 0} \frac{h/2-h^3/48+O(h^5)}{h/2}\cos x=\cos x$$ Here we have used $\sin z=x-z^3/3!+h^5/5!+...$

Z Ahmed
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    The question that the OP is asking is about the evaluation of the limit on the LHS of the last inequality in your post. – P. J. Mar 26 '21 at 12:41
  • @P.J.: Yes, but the OP's question is flawed -- see my comment there. This method is more straightforward. – TonyK Mar 26 '21 at 15:14
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    @TonyK I did read your comment. I hope you have read mine -- using Taylor series to expand $\sin x$ or $\cos h$ would be circular, since OP's kid is currently studying limits – P. J. Mar 26 '21 at 18:01