हिंदी

Limx→π482-(cosx+sinx)72-2sin2x is equal to ______.

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प्रश्न

`lim_(x rightarrow π/4) (8sqrt(2) - (cosx + sinx)^7)/(sqrt(2) - sqrt(2)sin2x)` is equal to ______.

विकल्प

  • 14

  • 7

  • `14sqrt(2)`

  • `7sqrt(2)`

MCQ
रिक्त स्थान भरें
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उत्तर

`lim_(x rightarrow π/4) (8sqrt(2) - (cosx + sinx)^7)/(sqrt(2) - sqrt(2)sin2x)` is equal to 14.

Explanation:

`lim_(x rightarrow π/4) (8sqrt(2) - (cosx + sinx)^7)/(sqrt(2) - sqrt(2)sin2x)` ...`["from" 0/0]`

∴ Apply L-hospital rule

`lim_(x rightarrow π/4) (-7(cosx + sinx)^6(cosx - sinx))/(-2sqrt(2)cos2x)` ...`["from" 0/0]`

`lim_(x rightarrow π/4) (7(cosx + sinx)^6(cosx - sinx))/(2sqrt(2)(cosx - sinx)(cosx + sinx))`   ...[∵ cos2x = cos2x – sin2x]

`lim_(x rightarrow π/4) (7(cosx + sinx)^6)/(2sqrt(2)(cosx + sinx))`  

= `(7(sqrt(2))^6)/(2sqrt(2)(sqrt(2))` 

= `(7 xx 8)/4`

= 14

shaalaa.com
Limits Using L-hospital's Rule
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