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Question : 66 of 160
Marks:
+1,
-0
Solution:
Let
L=‌| 2‌sin‌x−sin‌2‌x |
| x3 |
( form
‌) (using L-Hospital rule)
=‌| 2‌cos‌x−2‌cos‌2‌x |
| 3x2 |
( form
‌) =‌| −2‌sin‌x+4‌sin‌2‌x |
| 6x |
( form
‌) =‌| −2‌cos‌x+8‌cos‌2‌x |
| 6 |
=‌=1
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