lim. xtan2x-2xtanx x-0 (1-cos2x) 2 A)2 B)-2 C)1/2 D)-1/2

Question

lim. xtan2x-2xtanx x->0 (1-cos2x) 2 A)2 B)-2 C)1/2 D)-1/2

Aditya Gupta · Answer

1-cos2x=2sin^2x now, the above lim becomes L= l im xtan2x-2xtanx * x^4 4x^4 sin^4x but lim x^4/sin^4x as x tends to zero is 1 now, use series expansion lim (2x^2+8x^4/3+O(x^5) – 2x^2 – 2x^4/3 – O(x^5)/(x^4) =8/3 – 2/3= 2

Aditya Gupta · Answer

1-cos2x=2sin^2x now, the above lim becomes L= l im xtan2x-2xtanx * x^4 4x^4 sin^4x but lim x^4/sin^4x as x tends to zero is 1 now, use series expansion lim (2x^2+8x^4/3+O(x^5) – 2x^2 – 2x^4/3 – O(x^5)/(4x^4) =(8/3 – 2/3)/4 = 1/2

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lim. xtan2x-2xtanx x->0 (1-cos2x) 2 A)2 B)-2 C)1/2 D)-1/2

lim. xtan2x-2xtanx
x->0 (1-cos2x)²

A)2
B)-2
C)1/2
D)-1/2

2 Answers

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lim. xtan2x-2xtanx x->0 (1-cos2x) 2 A)2 B)-2 C)1/2 D)-1/2

lim. xtan2x-2xtanx x->0 (1-cos2x)2 A)2B)-2C)1/2D)-1/2

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lim. xtan2x-2xtanx
x->0 (1-cos2x)²

A)2
B)-2
C)1/2
D)-1/2