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$$\frac{d}{dx} \sin(x) = \lim_{h\to 0} \frac{\sin(x+h) - \sin(x)}{h}$$

$$\frac{d}{dx} \sin(x) = \lim_{h\to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$

$$\frac{d}{dx} \sin(x) = \sin(x)\lim_{h\to 0} \frac{\cos(h) -1}{h} + \cos(x)\lim_{h\to 0} \frac{\sin(h)}{h}$$

Normally I'd use L'Hopital's Rule here but considering that I'm trying to find the derivative in the first place, that kind of defeats the purpose.

Is there an easier way to approach these limits? I'm not seeing anything obvious.

user525966
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5 Answers5

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Better to use this

$$\sin(x+h)-\sin x=2\sin(h/2)\cos(h/2+x)$$

lab bhattacharjee
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Hint: You can use the squeeze theorem to show that $$\lim_{h\to0}\frac{\sin(h)}{h} = 1.$$ You can also show that $$\lim_{h\to0}\frac{\cos(h)-1}{h} = 0.$$ Try multiplying top and bottom by $\cos(h) + 1$. You will need to use the first limit.

Bryan Curtis
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  • I did consider this but wasn't able to get anywhere. For instance $-1 \leq \sin(h) \leq 1$ then $-1/h \leq \sin(h)/h \leq 1/h$ but I was still having a divide-by-$0$ issue with that so... – user525966 Feb 23 '18 at 05:12
  • You can show that $\cos(x) < \sin(x)/x<1/\cos(x)$. For the other limit consider multiplying top and bottom by $\cos(h) +1$. – Bryan Curtis Feb 23 '18 at 06:21
  • Ah clever use of squeeze theorem! For the other one though I am not sure. Multiplying the way you suggest gets me $(\cos(h)^2 - 1) / (h(\cos(h) + 1))$ and I don't know how to get rid of that $h$ on the bottom. – user525966 Feb 23 '18 at 06:53
  • Although from the Pythagorean identity it's also equal to $-\sin(h)^2 / (h(\cos(h)+1))$... hmmm – user525966 Feb 23 '18 at 06:57
  • I can rewrite as $-\frac{\frac{\sin(h)}{h} \sin(h)}{\cos(h)+1}$ which goes to I can rewrite as $-\frac{1 \cdot \sin(0)}{\cos(0)+1}$ or $-\frac{0}{2} = 0$. Is this a correct way to reach the result? – user525966 Feb 23 '18 at 07:01
  • @user525966: yes that's the way to go. – Paramanand Singh Feb 23 '18 at 07:29
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There is no algebraic approach to it. The two I know are:

  • use some geometric considerations on the unit circle to obtain the two limits by squeezing with obvious limits.

  • Define sine and cosine using their Taylor series. The derivatives are then obvious.

Martin Argerami
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You can try this:

$$\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}=\lim_{h\rightarrow0}\frac{\cos^2(\frac{h}{2})-\sin^2(\frac{h}{2})-\cos^2(\frac{h}{2})-\sin^2(\frac{h}{2})}{h}$$ $$=\lim_{h\rightarrow0}(-\frac{h}{2}*\frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2})=0$$

Sujit Bhattacharyya
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Before studying derivatives you are supposed to be somewhat familiar with limits (and continuity) and when you study these topics you will encounter the following theorem (with or without proof) $$\lim_{x\to 0}\frac{\sin x} {x} =1\tag{1}$$ The typical proof proceeds using the inequalities $$\sin x<x<\tan x$$ for $0<x<\pi/2$. An easy consequence of this is the limit $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{2}$$ and you should be able to deduce $(2)$ from $(1)$ easily using trigonometric identities. Now use the above results to finish your derivation.

Paramanand Singh
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