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Calculate $$\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{x-\frac{\pi}{2}}$$ by relating it to a value of $(\cos x)'$.

The answer is available here (pdf) at 1J-2.

However, I can't seem to make sense of what is actually being done here.

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    Will you be able to handle $$\lim_{t-\frac{\pi}{2}\to 0} \, \frac{\cos \left(x+t-\frac{\pi }{2}\right)-\cos (x)}{t-\frac{\pi }{2}}$$2012-08-23
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    Yes, that substitution makes sense. Thanks!2012-08-24

3 Answers 3

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Applying l'Hopital rule, you have

$$ \lim_{x\to\pi/2}\frac{\cos x}{x-\pi/2}=\lim_{x\to\pi/2}(\cos x)'=\left.(\cos x)'\right|_{x=\pi/2} $$ that in turn becomes $$\left.(\cos x)'\right|_{x=\pi/2}=\left.(-\sin x)\right|_{x=\pi/2}=-1$$

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    Too high-powered. One only needs to go back to how the derivative is defined...2012-08-23
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    @J.M. but probably simpler to think of for a beginner.2012-08-23
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    I don't know; last time I checked, one is usually taught the definition of the derivative **way before** one is taught l'Hôpital, but I guess you kids do things differently these days...2012-08-23
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    l´hopital does, indeed, still come later, I think. Thanks for both answers though!2012-08-24
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Use the definition of derivative at a point,

$$ f'(x_0) = \lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \,.$$ In your case $f(x)=\cos(x)$ and $x_0=\frac{\pi}{2}$.

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HINT : I think you wanna use the notation $y = x - \frac{\pi}{2}$.