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Limx→01-cos4θ1-cos6θ is ______.

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Question

`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.

Options

  • `4/9`

  • `1/2`

  • `(-1)/2`

  • –1

MCQ
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Solution

`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is `4/9`.

Explanation:

Given `lim_(theta -> 0) (1 - cos 4theta)/(1 - cos 6theta)`

= `lim_(theta -> 0) (2sin^2 2theta)/(2sin^2 3theta)`  .....`[because 1 - cos theta = 2 sin^2  theta/2]`

= `lim_(theta -> 0) (sin^2 2theta)/(sin^2 3theta)`

= `lim_(theta -> 0) [(sin 2 theta)/(sin 3theta)]^2`

= `lim_((theta -> 0),(2theta -> 0),(3theta -> 0)) [((sin 2theta)/(2theta) xx 2theta)/((sin 3theta)/(3theta) xx 3theta)]^2`

= `[(2theta)/(3theta)]^2`

= `(2/3)^2`

= `4/9`

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Chapter 13: Limits and Derivatives - Exercise [Page 242]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 58 | Page 242

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