As h approaches 0,
lim [cos((pi/2)+h) - cos(pi/2)] / h
The answer is -1. Can someone please explain how to arrive at this answer?
Are there special trig-based formulas I need to know to solve questions like this?
Two special limits you should know are
$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$ and $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x}=0$.
In this case, we have
$\displaystyle \lim_{h\to0}\frac{\cos\left(\frac{\pi}2+h\right)-\cos\frac{\pi}2}h$
$\displaystyle =\lim_{h\to0}\frac{\cos\frac\pi2\cos h-\sin\frac\pi2\sin h-0}h$ (sum and difference identity)
$\displaystyle =\lim_{h\to0}\frac{0\cdot\cos h-1\cdot\sin h}h$
$\displaystyle =\lim_{h\to0}\frac{-\sin h}h$
$\displaystyle =-\lim_{h\to0}\frac{\sin h}h$
$\displaystyle =-1$
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note the format of the limit ...
$\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
... this is the definition of $\displaystyle f'(x)$
in this case, $\displaystyle f(x) = \cos{x}$ and the limit is $\displaystyle f'\left(\frac{\pi}{2}\right) = -\sin \left(\frac{\pi}{2}\right) = -1$