If f(x) is continuous and f(92) = 29, then limx→0f(1 - cos(3x)/x2) is equal to:

Let the given limit be L. We have
L = limx→0 f(1 - cos(3x)/x2)
We know that 1 - cos(3x) = 2sin2(3x/2)
Therefore, 1 - cos(3x)/x2 = 2sin2(3x/2)/x2 = 2(sin(3x/2)/x)2
As x → 0, sin(3x/2)/x → 3/2
Therefore, limx→0 (1 - cos(3x)/x2) = 2(3/2)2 = 9/2
Since f(x) is continuous, we can write
L = limx→0 f(1 - cos(3x)/x2) = f(limx→0 (1 - cos(3x)/x2)) = f(9/2)
However, we are given that f(92) = 29. The problem statement does not provide enough information to determine f(9/2). We need more information about the function f(x) to evaluate the limit.
Let's use L'Hopital's rule to evaluate limx→0 (1 - cos(3x))/x2.
Applying L'Hopital's rule twice:
limx→0 (1 - cos(3x))/x2 = limx→0 (3sin(3x))/(2x) = limx→0 (9cos(3x))/2 = 9/2
Since f(x) is continuous, we have:
limx→0 f((1 - cos(3x))/x2) = f(limx→0 (1 - cos(3x))/x2) = f(9/2)
Without knowing the function f(x), we cannot determine the value of f(9/2). Therefore, we cannot find the limit.