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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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    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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    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 $y = x - \frac{\pi}{2}$.