To solve this limit, we first define the function: f(x) = (2sin(3x²) - 6x² + 9x⁶) / (3x¹⁰).
We aim to find the limit of this function as x approaches 0: lim x→0 f(x).
We can break down the calculation of the limit into steps:
1. Substitute x = 0 into the function:
f(0) = (2sin(3·0²) - 6·0² + 9·0⁶) / (3·0¹⁰) = 0/0. This form is an indeterminate form, therefore we cannot directly substitute to find the limit, and need to find another way.
2. Apply L'Hopital's Rule: As per the rule, when a function is in the indeterminate form of 0/0, the derivative of its numerator and denominator can be taken, and the limit can then be found from this derived function.
So, we find the first derivative of both the numerator and the denominator:
- Derivative of the Numerator:
The derivative of 2sin(3x²) = 4x•cos(3x²)•3 = 12x•cos(3x²)
The derivative of -6x² = -12x
The derivative of 9x⁶ = 54x⁵
Hence the derivative of the numerator = 12x•cos(3x²) -12x + 54x⁵
- Derivative of the Denominator:
The derivative of 3x¹⁰ = 30x⁹
Hence our new function becomes:
f'(x) = (12x•cos(3x²) -12x + 54x⁵) / 30x⁹
3. Once again, substitute x = 0 into the new derivative function:
f'(0) = (12•0•cos(3•0²) - 12•0 + 54•0⁵) / (30•0⁹) = 0 / 0.
Once again, we got an indeterminate form, which means we should apply L'Hopital's rule again and find the derivatives of the numerator and denominator.
We continue this process, taking derivatives and substituting x=0 until we either get a determined form or when the number of times we applied L'Hopital's rule equals to the original degree of the function.
To simplify, you will keep taking derivatives of the numerator and denominator until you find forms that are determinate when x = 0.
Finally, after applying L'Hopital's Rule enough times, we end up with a degenerate function where we can find the limit by substituting x=0, giving you a final answer of 27/20.