limx tends to 0 1-cos2x/x^2
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Answered by
2
here is your answer
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hopes helps you dear mate......
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Answered by
1
Answer:
2 .....
Explanation:
this question can be easily solved by applying l'hospital rule that is by differentiating both numerator and denominator until we are free with infinity by infinity form or zero by zero form
so we get answer as 2
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