Question
Mathematics Question on limits and derivatives
x→0lim xcosecx−cotx is
A
2−1
B
1
C
21
D
−1
Answer
21
Explanation
Solution
x→0lim xcosecx−cotx =x→0lim \left\\{\frac{\frac{1}{sin\,x}-\frac{cos\,x}{sin\,x}}{x} \right\\} =x→0lim (xsinx1−cosx) =x→0lim x×2sin(2x)cos(2x)1−1+2sin2(x/2) =x→0lim xcos(2x)sin(2x) =2x→0lim 2xtan2x×21 =21×1=1/2