t's consider the limit given by:lim𝑥→0−sin(2𝑥)𝑥3⋅1−cos(2𝑥)cos(2𝑥)lim x→0 − x 3 sin(2x) ⋅ cos(2x)1−cos(2x)
Question
t's consider the limit given by:lim𝑥→0−sin(2𝑥)𝑥3⋅1−cos(2𝑥)cos(2𝑥)lim x→0 − x 3 sin(2x) ⋅ cos(2x)1−cos(2x)
Solution
The given limit is:
lim (x→0) [ -sin(2x) / x³ ] * [ cos(2x) / 1 - cos(2x) ]
This limit can be solved by using L'Hopital's Rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.
First, let's rewrite the limit as a single fraction:
lim (x→0) [ -sin(2x) * cos(2x) ] / [ x³ * (1 - cos(2x)) ]
Now, let's find the derivative of the numerator and the denominator:
Derivative of the numerator: -2cos²(2x) + 2sin²(2x) Derivative of the denominator: 3x² + 2x³sin(2x)
So, the limit becomes:
lim (x→0) [ -2cos²(2x) + 2sin²(2x) ] / [ 3x² + 2x³sin(2x) ]
Now, we can apply L'Hopital's Rule again, and find the second derivative of the numerator and the denominator:
Second derivative of the numerator: -8cos(2x)sin(2x) + 8sin(2x)cos(2x) = 0 Second derivative of the denominator: 6x + 6x²sin(2x) + 4x³cos(2x)
So, the limit becomes:
lim (x→0) [ 0 ] / [ 6x + 6x²sin(2x) + 4x³cos(2x) ]
The limit of a constant (in this case, 0) divided by any function that doesn't approach infinity or negative infinity is 0. Therefore, the limit of the given expression as x approaches 0 is 0.
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