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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?

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  • $\begingroup$ You could do it this way. $\endgroup$ – David Mitra Feb 3 at 12:09
  • $\begingroup$ 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 $\endgroup$ – user2661923 Feb 3 at 12:13
  • $\begingroup$ 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. $\endgroup$ – user2661923 Feb 3 at 12:20
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    $\begingroup$ @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. $\endgroup$ – A-Level Student Feb 3 at 12:24
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    $\begingroup$ @A-LevelStudent +1: right-on $\endgroup$ – user2661923 Feb 3 at 12:56
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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 $\displaystyle\frac{\sin h}h$.

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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 $\cos(0)=1=\max_{x \in \mathbb{R}} \cos(x)$)

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    $\begingroup$ This seems like l'Hopital $\endgroup$ – Physor Feb 3 at 12:14
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    $\begingroup$ This is rather the definition of a derivative, in my opinion :) $\endgroup$ – TheSilverDoe Feb 3 at 12:14
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    $\begingroup$ Yes, I see but it amounts to using l'Hopital, in my opinion $\endgroup$ – Physor Feb 3 at 12:15
  • $\begingroup$ @TheSilverDoe: how do you show that $\cos'x=-\sin x$? $\endgroup$ – Martin Argerami Feb 3 at 12:17
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    $\begingroup$ Oh, thanks for the advice but I know what a derivative is. In this case both thing are the same, if you have noticed $\endgroup$ – Physor Feb 3 at 12:20

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