evaluate lim x=0 [(sin 3x cot 5x)/(xcot 4x)]
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
To evaluate the limit as x approaches 0 of [(sin 3x cot 5x)/(xcot 4x)], we can apply some trigonometric identities and simplification.
Step 1: Simplify the expression using the trigonometric identity cot(x) = cos(x)/sin(x).
[(sin 3x cot 5x)/(xcot 4x)] = [(sin 3x cos 5x)/(xsin 5x cos 4x)]
Step 2: Combine the terms with sin(3x) and sin(5x).
[(sin 3x cos 5x)/(xsin 5x cos 4x)] = [(sin 3x cos 5x)/(xsin 5x cos 4x)] * (1/cos 5x)
= [(sin 3x)/(xsin 5x cos 4x)]
Step 3: Simplify further by using the trigonometric identity sin(a)/a = 1.
[(sin 3x)/(xsin 5x cos 4x)] = [(sin 3x)/(3x)] * [(5x)/(sin 5x)] * [(1/cos 4x)]
Step 4: Evaluate the individual limits as x approaches 0.
lim [(sin 3x)/(3x)] as x approaches 0 is equal to 1.
lim [(5x)/(sin 5x)] as x approaches 0 is equal to 1.
lim [(1/cos 4x)] as x approaches 0 is equal to 1.
Step 5: Multiply the limits obtained in Step 4.
1 * 1 * 1 = 1
Therefore, the limit as x approaches 0 of [(sin 3x cot 5x)/(xcot 4x)] is equal to 1.