anonymous
  • anonymous
Can anyone help me with 1J-2? Not entirely sure where to start...
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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phi
  • phi
They are saying it is easy to find the derivative of cos(x) and evaluate it at x= pi/2 \[ \cos'(x)\bigg|_{x= \frac{\pi}{2} }= ?\] The other idea is that the expression \[ \lim_{x\rightarrow \frac{\pi}{2} } \frac{\cos(x)}{x-\frac{\pi}{2}}\] is vaguely reminiscent of the definition of the derivative \[ f'(x) = \lim_{\Delta\rightarrow 0} \frac{ f(x+\Delta)- f(x)}{\Delta}\] Now if you can show that the given expression can be re-written to look like the definition of the derivative, then you can claim that it and the derivative (evaluated at pi/2) have the same value (-1 in this case) I would begin by letting \( x = \frac{\pi}{2} + \Delta \) in the given expression.

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