\displaystyle\lim{h\to 0}\dfrac{\sin(\pi/2 + h) - 1}{h} equals

0 This is f'(\pi/2) for f(x) = \sin x. f'(x) = \cos x; \cos(\pi/2) = 0.

Defining the Derivative and Using Derivative Notation

AP Calculus AB· difficulty 3/5

\displaystyle\lim_{h\to 0}\dfrac{\sin(\pi/2 + h) - 1}{h}

Explanation

f'(\pi/2)

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