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Grade:
        
lim.    xtan2x-2xtanx  
 x->0   (1-cos2x)2
 
A)2
B)-2
C)1/2
D)-1/2
    
one year ago

Answers : (2)

Aditya Gupta
1717 Points
							
1-cos2x=2sin^2x
now, the above lim becomes
L= lim 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
one year ago
Aditya Gupta
1717 Points
							
 
1-cos2x=2sin^2x
now, the above lim becomes
L= lim 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
one year ago
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