Evaluate the following limits :
`Lim_(x to 9 ) (x^(3//2)-27)/(x-9)`

Answer

To evaluate the limit

\[ \lim_{x \to 9} \frac{x^{3/2} - 27}{x - 9}, \]

we can follow these steps:

### Step 1: Substitute the value of \(x\)

First, we will substitute \(x = 9\) directly into the expression:

\[ \frac{9^{3/2} - 27}{9 - 9}. \]

Calculating \(9^{3/2}\):

\[ 9^{3/2} = (9^{1/2})^3 = 3^3 = 27. \]

So, we have:

\[ \frac{27 - 27}{9 - 9} = \frac{0}{0}. \]

This is an indeterminate form, so we need to apply L'Hôpital's Rule or simplify the expression.

### Step 2: Apply L'Hôpital's Rule

Since we have the form \(0/0\), we can differentiate the numerator and denominator:

- The numerator: \(f(x) = x^{3/2} - 27\)
- The denominator: \(g(x) = x - 9\)

Differentiating both:

\[ f'(x) = \frac{3}{2} x^{1/2}, \]
\[ g'(x) = 1. \]

### Step 3: Rewrite the limit using derivatives

Now, we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to 9} \frac{f'(x)}{g'(x)} = \lim_{x \to 9} \frac{\frac{3}{2} x^{1/2}}{1}. \]

### Step 4: Substitute \(x = 9\) again

Now substitute \(x = 9\):

\[ \frac{3}{2} \cdot 9^{1/2} = \frac{3}{2} \cdot 3 = \frac{9}{2}. \]

### Final Answer

Thus, the limit evaluates to:

\[ \lim_{x \to 9} \frac{x^{3/2} - 27}{x - 9} = \frac{9}{2}. \]