Evaluate `lim_(xto7//2) (2x^(2)-9x+8)^(cot(2x-7)).`

Options

A. both `underset(xtoa)limf(x)` and `underset(xtoa)limg(x)` must exist
B. `underset(xtoa)limf(x)` need not exist but `underset(xtoa)limg(x)` exists
C. neither `underset(xtoa)limf(x)` nor `underset(xtoa)limg(x)` may exist
D. `underset(xtoa)limf(x)` exists but `underset(xtoa)limg(x)` need not exist

Explanation

To evaluate the limit \( \lim_{x \to \frac{7}{2}} (2x^2 - 9x + 8)^{\cot(2x - 7)} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{7}{2} \)

First, we substitute \( x = \frac{7}{2} \) into the expression \( 2x^2 - 9x + 8 \):

\[ 2 \left(\frac{7}{2}\right)^2 - 9 \left(\frac{7}{2}\right) + 8 \]

Calculating each term:

\[ 2 \cdot \frac{49}{4} - \frac{63}{2} + 8 = \frac{49}{2} - \frac{63}{2} + \frac{16}{2} = \frac{49 - 63 + 16}{2} = \frac{2}{2} = 1 \]

So, we have:

\[ 2x^2 - 9x + 8 \to 1 \text{ as } x \to \frac{7}{2} \]

### Step 2: Evaluate \( \cot(2x - 7) \)

Next, we evaluate \( \cot(2x - 7) \) as \( x \to \frac{7}{2} \):

\[ 2x - 7 \to 2 \cdot \frac{7}{2} - 7 = 0 \]

Thus, \( \cot(2x - 7) \to \cot(0) \), which is undefined and approaches infinity.

### Step 3: Identify the form of the limit

We have \( 1^{\infty} \) form, which is an indeterminate form. We can use the exponential limit property:

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)} \]

Here, \( f(x) = 2x^2 - 9x + 8 \) and \( g(x) = \cot(2x - 7) \).

### Step 4: Calculate \( g(x) \cdot (f(x) - 1) \)

We need to calculate:

\[ \lim_{x \to \frac{7}{2}} \cot(2x - 7) \cdot ((2x^2 - 9x + 8) - 1) \]

We already found that \( (2x^2 - 9x + 8) - 1 \to 0 \) as \( x \to \frac{7}{2} \).

### Step 5: Rewrite \( \cot(2x - 7) \)

Using the identity \( \cot(x) = \frac{\cos(x)}{\sin(x)} \), we rewrite:

\[ \cot(2x - 7) = \frac{\cos(2x - 7)}{\sin(2x - 7)} \]

As \( x \to \frac{7}{2} \), both \( \cos(2x - 7) \) approaches \( \cos(0) = 1 \) and \( \sin(2x - 7) \) approaches \( 0 \).

### Step 6: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

\[ \lim_{x \to \frac{7}{2}} \frac{\cos(2x - 7)}{\sin(2x - 7)} = \lim_{x \to \frac{7}{2}} \frac{-2\sin(2x - 7)}{2\cos(2x - 7)} = \lim_{x \to \frac{7}{2}} -\tan(2x - 7) \]

As \( x \to \frac{7}{2} \), \( -\tan(0) = 0 \).

### Step 7: Combine results

Thus, we have:

\[ \lim_{x \to \frac{7}{2}} \cot(2x - 7) \cdot ((2x^2 - 9x + 8) - 1) = \lim_{x \to \frac{7}{2}} \frac{\cos(2x - 7)}{\sin(2x - 7)} \cdot 0 = 0 \]

### Step 8: Final result

Now substituting back into the exponential limit:

\[ \lim_{x \to \frac{7}{2}} (2x^2 - 9x + 8)^{\cot(2x - 7)} = e^{0} = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]