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Question

What is the value of limxπ2[cosx] ?

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Solution

Let h be a very small positive number.
limxπ2[cos(π2h)][sinh]=0limxπ2[cosπ2]=0limxπ2[cos(π2+h)][cos(π2(h))][sin(h)]=1limxπ2+limxπ2Notcontinuous.

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