Question 21. lim (cosec x cot x)xâ0

1.	Question 21. lim (cosec x cot x)xâ0
Answer» Explanation : lim x -> 0 (cosecx - cotx) = lim x->0 (1/sinx - cosx/sinx) = lim x->0 (1-cosx)/sinx If we put x = 0 we will get 0/0 form so using L'hospital rule = lim x-> 9 (sinx)/(cosx) = sin0/cos0 = 0/1 = 0 Solution : 0 If you find this answer helpful then like it.

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Explanation : lim x -> 0 (cosecx - cotx) = lim x->0 (1/sinx - cosx/sinx) = lim x->0 (1-cosx)/sinx If we put x = 0 we will get 0/0 form so using L'hospital rule = lim x-> 9 (sinx)/(cosx) = sin0/cos0 = 0/1 = 0

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Question 21. lim (cosec x cot x)xâ0

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Question 21. lim (cosec x cot x)xâ0