lim_(x->0)(5x+tan(3x))/(2x-tan(5x)) = ?

1 Answer
Dec 7, 2016

-8/3

Explanation:

(5x+sin(3x)/cos(3x))/(2x-sin(5x)/cos(5x))=(5xcos(3x)+sin(3x))/(2xcos(5x)-sin(5x)) (cos(5x)/cos(3x))=

((5x)/(2x))((cos(3x)+sin(3x)/(5x))/(cos(5x)-sin(5x)/(2x))) (cos(5x)/cos(3x))=

(5/2)((cos(3x)+(5/3)(3/5)sin(3x)/(5x))/(cos(5x)-(2/5)(5/2)sin(5x)/(2x)))(cos(5x)/cos(3x)) =

(5/2)((cos(3x)+(3/5)sin(3x)/(3x))/(cos(5x)-(5/2)sin(5x)/(5x))) (cos(5x)/cos(3x))

Now

lim_(x->0)(5x+tan(3x))/(2x-tan(5x)) =5/2lim_(x->0)((cos(3x)+(3/5)sin(3x)/(3x))/(cos(5x)-(5/2)sin(5x)/(5x)))(lim_(x->0)(cos(5x)/cos(3x))) = (5/2)((1+3/5)/(1-5/2))(1) = -8/3