l´ımx−→0ln(cos(3x))ln(cos(2x))

The expression you provided seems to be a limit problem in calculus. However, it seems like there might be a formatting issue as the limit is not specified. If you're looking for the limit as x approaches 0 for the expression ln(cos(3x))/ln(cos(2x)), here's how you can solve it:

Step 1: Recognize the indeterminate form As x approaches 0, both the numerator and the denominator approach ln(1), which is 0. So, we have an indeterminate form of 0/0.

Step 2: Apply L'Hopital's Rule L'Hopital's Rule 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. So, we differentiate the numerator and the denominator:

Derivative of ln(cos(3x)) with respect to x is -3sin(3x)/cos(3x). Derivative of ln(cos(2x)) with respect to x is -2sin(2x)/cos(2x).

Step 3: Take the limit of the quotient of the derivatives Now, we find the limit as x approaches 0 for the expression -3sin(3x)/cos(3x) divided by -2sin(2x)/cos(2x). This simplifies to 3/2 * (sin(3x)/sin(2x)) * (cos(2x)/cos(3x)).

Step 4: Evaluate the limit As x approaches 0, sin(3x)/sin(2x) approaches 3/2 and cos(2x)/cos(3x) approaches 2/3. So, the limit of the original expression as x approaches 0 is (3/2) * (2/3) = 1.

So, l´ımx−→0ln(cos(3x))/ln(cos(2x)) = 1.