Find the limit. Use I\'Hospital\'s Rule if appropriate. Use INF to represent pos

Question

Find the limit. Use I'Hospital's Rule if appropriate. Use INF to represent positive infinity, NINF for negative infinity, and D for the limit does not exist. lim_x rightarrow 0 -5 cot (9x) sin (3x) =

Explanation / Answer

answer is -5/3 Using cot(9 x) = (cos(9 x))/(sin(9 x)), write -5 sin(3 x) cot(9 x) as -(5 sin(3 x) cos(9 x))/(sin(9 x)): lim_(x->0) -(5 cos(9 x) sin(3 x))/(sin(9 x)) By the product rule, lim_(x->0) -(5 sin(3 x) cos(9 x))/(sin(9 x)) = -5 (lim_(x->0) (sin(3 x))/(sin(9 x))) (lim_(x->0) cos(9 x)): -5 lim_(x->0) cos(9 x) lim_(x->0) (sin(3 x))/(sin(9 x)) lim_(x->0) cos(9 x) = cos(9 0) = 1: -5 lim_(x->0) (sin(3 x))/(sin(9 x)) Applying l'Hôpital's rule, we get that lim_(x->0) (sin(3 x))/(sin(9 x)) | = | lim_(x->0) ( d/( dx) sin(3 x))/( d/( dx) sin(9 x)) | = | lim_(x->0) (3 cos(3 x))/(9 cos(9 x)) | = | lim_(x->0) (cos(3 x))/(3 cos(9 x)) -5lim_(x->0) (cos(3 x))/(3 cos(9 x)) lim_(x->0) (cos(3 x))/(3 cos(9 x)) = (cos(3 0))/(3 cos(9 0)) = 1/3: -51/3 -5/3 = -5/3: Answer: -5/3

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