Evaluate: `("lim")_(xvec7/2)(2x^2-9x+8)^(cot(2x-7))`

Options

A. `e^(5//2)`
B. `e^(-5//2)`
C. `e^(7//2)`
D. `e^(3//2)`

Answer

To evaluate the limit

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

we can follow these steps:

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

First, we substitute \( x = \frac{7}{2} \) into the expression:

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

Calculating this gives:

\[ 2 \cdot \frac{49}{4} - \frac{63}{2} + 8 = \frac{49}{2} - \frac{63}{2} + \frac{32}{4} = \frac{49 - 63 + 32}{2} = \frac{18}{2} = 9. \]

Now, substituting into \( \cot(2x - 7) \):

\[ \cot(2 \cdot \frac{7}{2} - 7) = \cot(7 - 7) = \cot(0), \]

which is undefined. Thus, we have the form \( 9^{\text{undefined}} \), which is indeterminate.

### Step 2: Identify the Indeterminate Form

We recognize that as \( x \to \frac{7}{2} \), the base \( (2x^2 - 9x + 8) \) approaches 1, and \( \cot(2x - 7) \) approaches infinity. This gives us the indeterminate form \( 1^{\infty} \).

### Step 3: Apply the Standard Limit Formula

For limits of the form \( \lim_{x \to a} f(x)^{g(x)} \) that yield \( 1^{\infty} \), we can use the formula:

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

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

### Step 4: Rewrite the Limit

We rewrite the limit as:

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

This simplifies to:

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

### Step 5: Evaluate the Limit

Now, we need to evaluate:

\[ \lim_{x \to \frac{7}{2}} \frac{2x^2 - 9x + 7}{\tan(2x - 7)}. \]

Both the numerator and denominator approach 0 as \( x \to \frac{7}{2} \), giving us the \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule.

### Step 6: Differentiate Numerator and Denominator

Differentiate the numerator and denominator:

- The derivative of the numerator \( 2x^2 - 9x + 7 \) is \( 4x - 9 \).
- The derivative of \( \tan(2x - 7) \) is \( 2\sec^2(2x - 7) \).

Thus, we have:

\[ \lim_{x \to \frac{7}{2}} \frac{4x - 9}{2\sec^2(2x - 7)}. \]

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

Now substituting \( x = \frac{7}{2} \):

\[ 4\left(\frac{7}{2}\right) - 9 = 14 - 9 = 5, \]

and

\[ 2\sec^2(0) = 2 \cdot 1 = 2. \]

So we have:

\[ \lim_{x \to \frac{7}{2}} \frac{5}{2} = \frac{5}{2}. \]

### Step 8: Final Limit Calculation

Now we can substitute this back into our exponential limit:

\[ e^{\frac{5}{2}}. \]

Thus, the final answer is:

\[ \lim_{x \to \frac{7}{2}} (2x^2 - 9x + 8)^{\cot(2x - 7)} = e^{\frac{5}{2}}. \]