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While deriving $\frac{\rm{d}}{\rm{dx}}\rm{sin}(x)$, using the definition of the derivative and expanding $\rm{sin}(x+h)$ leads to

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

The second limit can be evaluated by applying the squeeze theorem, and if I wasn't going from first principles I could use l'Hopital's rule for the first limit, but that would be circular logic.

How can I evaluate $\lim_{h\to 0} \frac{cos(h)-1}{h}$ without l'Hopital?

Henry
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spraff
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  • You could do it this way. – David Mitra Feb 03 '21 at 12:09
  • Actually, you have opened up a can of worms, that can not be explored in Analytical Geometry but must instead be explored in Real Analysis. In "Calculus" 2nd Edition 1966 Vol. 1 (Tom Apostol) one of the axioms that the reader is obliged to accept is that $\lim_{x \to 0}\frac{\sin x}{x} = 1.$ Apostol then makes the geometric argument that the traditional definition of the sine and cosine functions fit his axioms (including the one I am referring to), as long as you change the domain of the sine and cosine functions to dimensionless real numbers. ...see next comment – user2661923 Feb 03 '21 at 12:13
  • Apostol then (very casually) mentions that the sine and cosine functions can alternatively be defined by the pertinent Taylor series, and then all of the normal consequences can then be proved. Using this approach, the result that you are seeking is (again) immediate. So the question of how to prove the result depends on the axiomatic Real Analysis definitions of the sine and cosine functions. What Apostol does not do, is define the sine and cosine functions in terms of points on the unit circle. ...see next comment. – user2661923 Feb 03 '21 at 12:20
  • Instead, Apostol uses his (? semi-arbitrary ?) axioms to establish that the sine and cosine functions defined in terms of points on the unit circle satisfy his axioms. – user2661923 Feb 03 '21 at 12:22
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    @Somos In my opinion I don't think this question should have been seen as a duplicate, as the other linked question only has arguments that use the result $\lim_{x\to0}\frac{\sin x}{x}=1$, whereas this question could have attracted other answers, which could've used different arguments. – A-Level Student Feb 03 '21 at 12:24
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    @A-LevelStudent +1: right-on – user2661923 Feb 03 '21 at 12:56
  • @José Carlos Santos please see my abive comment. – A-Level Student Feb 03 '21 at 13:37

2 Answers2

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You can do $$ \frac{\cos h-1}h=\frac{\cos^2h-1}{h(\cos h+1)}=-\sin h \,\frac{\sin h}h\,\frac1{\cos h+1}. $$ The nontrivial one is always $\displaystyle\frac{\sin h}h$.

Martin Argerami
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$$\frac{\cos(h)-1}{h}=\frac{\cos(h)-\cos(0)}{h-0} \longrightarrow \cos'(0)=0$$

(if you don't know yet that $\cos'=-\sin$, you can still get that $\cos'(0)=0$ by noticing that $\cos$ has a local maximum at $x=0$, since $\cos(0)=1=\max_{x \in \mathbb{R}} \cos(x)$)

TheSilverDoe
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    This seems like l'Hopital – Physor Feb 03 '21 at 12:14
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    This is rather the definition of a derivative, in my opinion :) – TheSilverDoe Feb 03 '21 at 12:14
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    Yes, I see but it amounts to using l'Hopital, in my opinion – Physor Feb 03 '21 at 12:15
  • @TheSilverDoe: how do you show that $\cos'x=-\sin x$? – Martin Argerami Feb 03 '21 at 12:17
  • @Physor I would not say so. L'Hopital's rule is useful when you have a quotient which is not directly a rate of change. Of course you can see that as a special case of L'Hopital rule, but to prove l'Hopital, you must know what a derivative is, so... – TheSilverDoe Feb 03 '21 at 12:17
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    Oh, thanks for the advice but I know what a derivative is. In this case both thing are the same, if you have noticed – Physor Feb 03 '21 at 12:20
  • @MartinArgerami It depends on your definition of $\cos$ and $\sin$... – TheSilverDoe Feb 03 '21 at 12:23
  • Ok. And what definition of sine and cosine gives you $\cos'x=-\sin x$ but not $\sin 'x=\cos x$? – Martin Argerami Feb 03 '21 at 12:27
  • @Physor Sorry if I have been unclear, I just wanted to say that L'Hopital's rule is far more general than the simple definition of a derivative, which is what I use, and it comes after become you need to know what a derivative is to get l'Hopital's rule. That's why I think that my answer is more elementary than an application of L'Hopital's rule... but it is just my opinion. – TheSilverDoe Feb 03 '21 at 12:28
  • @MartinArgerami You got a point here, I did not see that the OP wanted originally to calculate $\sin'$. – TheSilverDoe Feb 03 '21 at 12:29
  • @TheSilverDoe perhaps others would understand why this isn't circular if you use the cosine graph: This is probably the easiest way of looking at it. Let $f(x)=\cos x$. $$f'(0)=\lim_{h\to 0}\frac{\cos(0+h)-\cos 0}{h}=\lim_{h\to 0}\frac{\cos(h)-1}{h}$$ But we know that $f'(0)=0$, as the cosine graph has a turning point at $x=0$, so we have $$\lim_{h\to 0}\frac{\cos(h)-1}{h}=0$$ – A-Level Student Feb 03 '21 at 12:29
  • @A-LevelStudent Yes, indeed, the fact that $\cos'(0)=0$ can be seen without even knowing that $\cos'=-\sin$, just seeing that $\cos$ has a local maximum at $x=0$ (which can be seen on the trigonometric circle). – TheSilverDoe Feb 03 '21 at 12:31
  • @TheSilverDoe Exactly, hopefully that will clarify things. Perhaps you should add that into your answer? – A-Level Student Feb 03 '21 at 12:32
  • @A-LevelStudent Yes, thanks, I edited. – TheSilverDoe Feb 03 '21 at 12:34
  • @MartinArgerami Maybe you will find the edit more clear, I am not using anymore that $\cos'=-\sin$. The only thing I admit here is that $\cos$ is differentiable, but I guess the OP is ok with that. – TheSilverDoe Feb 03 '21 at 12:43