AP Calculus AB Exam Question:

Find the limit algebraically:

Step-by-Step Solution:

To find the limit as $x$ approaches $\frac{\pi}{2}$, we can use algebraic techniques such as factoring and simplifying.

Let's start by simplifying the expression:

Recall the identity $cos(x-1) = cos(x)cos(1) + sin(x)sin(1)$. Using this, the expression becomes:

Next, observe the denominator $\sin^2(2x-2)$. We know that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. Applying this identity, we have:

Simplifying further, we get:

Now, we need to analyze the numerator and the denominator separately.

Let's evaluate the numerator first. For the limit as $x$ approaches $\frac{\pi}{2}$:

Using the limit properties, we can evaluate each term separately. Recall that $\lim_{x \to \frac{\pi}{2}} \cos(x) = 0$ and $\lim_{x \to \frac{\pi}{2}} \sin(x) = 1$. Plugging these values into the expression, we get:

Next, let's evaluate the denominator. For the limit as $x$ approaches $\frac{\pi}{2}$:

Using the limit properties, we can evaluate each term separately. Recall that $\lim_{x \to \frac{\pi}{2}} \cos(x-1) = 0$ and $\lim_{x \to \frac{\pi}{2}} \sin(x-1) = 1$. Plugging these values into the expression, we get:

Finally, we can put the evaluated numerator and denominator together to find the overall limit.

However, since the denominator evaluated to zero, we need to further analyze the behavior around $\frac{\pi}{2}$. For this, we can examine the behavior of the original numerator and denominator.

Looking at the original expression, we know that $\sin^2(2x-2)$ approaches zero as $x$ approaches $\frac{\pi}{2}$, while $\cos(x-1)$ does not. Therefore, the limit of the expression does not exist algebraically as it is undefined.

Hence, the final answer is: