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Evaluate the following limits: limx→01-cos2xxsin2x - Mathematics

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प्रश्न

Evaluate the following limits:

`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`

बेरीज
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उत्तर

We know `lim_(x -> 0) (sin x)/x` = 1

`lim_(x -> 0) (1 - cos^2x)/(x sin2x) =  lim_(x -> 0) (sin^2x)/(x xx 2sinx cosx)`

= `1/2 lim_(x -> 0) (sin)/(xcosx)` 

= `1/2 (lim_(x -> 0) (sinx)/x) (lim_(x -> 0) 1/cosx)`

= `1/2 xx 1 xx 1/cos0`

`lim_(x -> 0) (1 - cos^2x)/(x sin2x) = 1/2 xx 1/1`

= `1/2`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 17 | पृष्ठ ११८

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