Find if the function $x\mapsto |\sin (x)-1|$ is differentiable at $x=\pi /2$ .
I get stuck at $$\lim_{h \rightarrow 0}{ \left|{\cos h \over h}\right|}$$
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5We're talking $\mathbb{R}$ here, aren't we? Write $\lvert\sin x - 1\rvert$ in a way that doesn't use an absolute value. – Daniel Fischer Jul 06 '13 at 16:28
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How did you get $\cos h/h$? – Berci Jul 06 '13 at 16:31
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To add to @DanielFischer's comment, many times the difficulty with absolute value problems comes from a basic inability to translate the symbols into their definitions. – The Chaz 2.0 Jul 06 '13 at 16:32
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@TheChaz2.0 Hey, do you have some link that can help me understand it better ? – Aman Mittal Jul 06 '13 at 16:33
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Aman, maybe this M.SE question would be a good starting point...? – The Chaz 2.0 Jul 06 '13 at 16:49
2 Answers
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HINT:
With Graham Hesketh's solution, the differentiability of $(1-\sin x)$ is established here
As $$\frac{d f(x)}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$
$$\implies \frac{d f(x)}{dx}_{(\text{at }x=a)}=\lim_{h\to0}\frac{f(a+h)-f(a)}h$$
Here $f(x)=1-\sin x, a=\frac\pi2$
So, $$\frac{d(1-\sin x)}{dx}_{(\text{at }x=\frac\pi2)}=\lim_{h\to0}\frac{1-\sin (\frac\pi2+h)-(1-\sin\frac\pi2) }h=\lim_{h\to0}\frac{1-\cos h}h$$ which is proved here
lab bhattacharjee
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Thanks Again, 1 side question though, if i just have to find the limit of $cosh \over h$ as $h \rightarrow 0$ how do i approach it ? The answer is "doesn;t exist" right ? – Aman Mittal Jul 06 '13 at 16:54
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@AmanMittal, my pleasure. As $\cos 0=1, \lim_{h\to0^+}\frac{\cos h}h=+\infty, \lim_{h\to0^-}\frac{\cos h}h=-\infty$ – lab bhattacharjee Jul 06 '13 at 16:55
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$\sin(x)\le 1\Rightarrow |\sin(x)-1|=1-\sin(x)$
so this example is differentiable everywhere.
[see lab bhattacharjee's solution for the differentiability of $1-\sin(x)$ ]
:)
Graham Hesketh
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