Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it.lim 𝜃→𝜋/2 1 − sin(𝜃)1 + cos(6𝜃)

To find the limit of the given function as 𝜃 approaches 𝜋/2, we can use l'Hospital's Rule. l'Hospital'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, provided the limit of the original function is of the form 0/0 or ∞/∞.

The given function is (1 - sin(𝜃))/(1 + cos(6𝜃)). As 𝜃 approaches 𝜋/2, both the numerator and the denominator approach 0, so we can apply l'Hospital's Rule.

First, we need to find the derivatives of the numerator and the denominator.

The derivative of 1 - sin(𝜃) with respect to 𝜃 is -cos(𝜃).

The derivative of 1 + cos(6𝜃) with respect to 𝜃 is -6sin(6𝜃).

So, the limit of the original function as 𝜃 approaches 𝜋/2 is equal to the limit of the quotient of these derivatives as 𝜃 approaches 𝜋/2:

lim 𝜃→𝜋/2 (-cos(𝜃)/-6sin(6𝜃))

This simplifies to:

lim 𝜃→𝜋/2 (cos(𝜃)/6sin(6𝜃))

As 𝜃 approaches 𝜋/2, cos(𝜃) approaches 0 and sin(6𝜃) approaches sin(3𝜋) = 0. So, the limit of the function as 𝜃 approaches 𝜋/2 is 0/0, which is undefined.

Therefore, the limit of the given function as 𝜃 approaches 𝜋/2 does not exist.

To find the limit of the given function as 𝜃 approaches 𝜋/2, we can use l'Hospital's Rule. This 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, provided the limit of the original function is of the form 0/0 or ∞/∞.

The function is given as (1 - sin(𝜃))/(1 + cos(6𝜃)).

First, let's check the limit of the numerator and the denominator separately as 𝜃 approaches 𝜋/2.

lim 𝜃→𝜋/2 (1 - sin(𝜃)) = 1 - sin(𝜋/2) = 1 - 1 = 0 lim 𝜃→𝜋/2 (1 + cos(6𝜃)) = 1 + cos(3𝜋) = 1 - 1 = 0

Since both the numerator and the denominator approach 0, we can apply l'Hospital's Rule.

Now, let's find the derivatives of the numerator and the denominator.

The derivative of 1 - sin(𝜃) with respect to 𝜃 is -cos(𝜃). The derivative of 1 + cos(6𝜃) with respect to 𝜃 is -6sin(6𝜃).

So, the limit of the given function as 𝜃 approaches 𝜋/2 is equal to the limit of the quotient of these derivatives as 𝜃 approaches 𝜋/2.

lim 𝜃→𝜋/2 (-cos(𝜃)/-6sin(6𝜃)) = lim 𝜃→𝜋/2 (cos(𝜃)/6sin(6𝜃))

Now, let's find this limit.

lim 𝜃→𝜋/2 (cos(𝜃)/6sin(6𝜃)) = cos(𝜋/2)/6sin(3𝜋) = 0/0

Again, we have a 0/0 form, so we can apply l'Hospital's Rule again.

The derivative of cos(𝜃) with respect to 𝜃 is -sin(𝜃). The derivative of 6sin(6𝜃) with respect to 𝜃 is 36cos(6𝜃).

So, the limit of the given function as 𝜃 approaches 𝜋/2 is equal to the limit of the quotient of these derivatives as 𝜃 approaches 𝜋/2.

lim 𝜃→𝜋/2 (-sin(𝜃)/36cos(6𝜃)) = lim 𝜃→𝜋/2 (-sin(𝜃)/36cos(6𝜃))

Now, let's find this limit.

lim 𝜃→𝜋/2 (-sin(𝜃)/36cos(6𝜃)) = -sin(𝜋/2)/36cos(3𝜋) = -1/36

So, the limit of the given function as 𝜃 approaches 𝜋/2 is -1/36.