Final answer:
The limit of sin(x)/(2x×tan(3x)) as x approaches 0 is found using L'Hôpital's rule and simplifying the expression, resulting in a final answer of 1/6.
Step-by-step explanation:
The question involves finding the limit of the function sin(x)/(2x×tan(3x)) as x approaches 0. To solve this, we can utilize L’Hôpital's rule because the expression is of the indeterminate form 0/0 as x approaches 0. L’Hôpital's rule states that if the limit of functions f(x) and g(x) as x approaches a value c are both 0 or both ∞, the limit of f(x)/g(x) as x approaches c is the same as the limit of the derivatives of f(x) and g(x), that is, the limit of f'(x)/g'(x) as x approaches c.
First, let's find the derivative of the numerator, sin(x), which is cos(x). The derivative of 2x is 2. The derivative of tan(3x) is 3sec²(3x), by the chain rule since the derivative of tan(x) is sec²(x). Thus, using L’Hôpital's rule, we take the limit of the new function as x approaches 0:
lim (x→0) [cos(x) / (2 × 3 × sec²(3x))]
We then plug in 0 for x to evaluate this limit:
cos(0) / (2 × 3 × sec²(0))
Since cos(0) = 1 and sec(0) = 1, the expression simplifies to:
1 / (2 × 3 × 1²)
The final answer is therefore 1/6, which corresponds to option A.