t's consider the limit given by:lim⁡𝑥→0−sin⁡(2𝑥)𝑥3⋅1−cos⁡(2𝑥)cos⁡(2𝑥)lim x→0 − ​ x 3 sin(2x)​ ⋅ cos(2x)1−cos(2x)​

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.