We can evaluate the limit by applying L'Hopital's rule:
lim x→0 [(sin 3x cot 5x)/(xcot 4x)]
= lim x→0 (cos 3x * csc^2 5x - 3sin^2 3x * csc 5x * cot 5x)/(cot 4x - 4x csc^2 4x)
Plugging in x=0 directly gives an indeterminate form of 0/0. Taking the derivatives and evaluating again:
= lim x→0 (-3sin 3x * cos 3x * csc^2 5x + cos^2 3x * (-5csc^2 5x * cot 5x) )/(-4csc^2 4x - 4x*(-2csc 4x * cot 4x *(-csc^2 4x)))
= lim x→0 [-3sin 3x * cos 3x * csc^2 5x - 5cos^3 3x * csc^2 5x]/[4csc^2 4x + 8xcsc 4x * cot 4x * csc^2 4x]
Plugging in x=0 directly again yields an indeterminate form of 0/0. Taking derivatives and evaluating again:
= lim x→0 [-9cos^2 3x * csc^2 5x + 30cos^2 3x * csc^4 5x + 15sin 3x * cos^2 3x * csc^4 5x]/[-4csc^4 4x - 8csc 4x * cot 4x * csc^4 4x - 8xcsc 4x *(-csc^2 4x) *(-2csc 4x * cot^2 4x + 1)]
Plugging in x=0 directly gives the final answer of:
= -27/8
evaluate lim x=0 [(sin 3x cot 5x)/(xcot 4x)]
1 answer