Answer :

Alright! Let's break down the step-by-step solution for evaluating the limit:

Given the function:

[tex]\[ f(x) = \frac{x^3 - 9x}{x^2 - 2x - 3} \][/tex]

We want to find the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 3. That is:

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

### Step 1: Factorize the numerator and the denominator

First, let's factorize both the numerator [tex]\( x^3 - 9x \)[/tex] and the denominator [tex]\( x^2 - 2x - 3 \)[/tex].

- For the numerator [tex]\( x^3 - 9x \)[/tex]:

[tex]\[ x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3) \][/tex]

- For the denominator [tex]\( x^2 - 2x - 3 \)[/tex]:

[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) \][/tex]

So the function now looks like:

[tex]\[ f(x) = \frac{x(x - 3)(x + 3)}{(x - 3)(x + 1)} \][/tex]

### Step 2: Simplify the expression

Notice that both the numerator and denominator have a common factor of [tex]\( (x - 3) \)[/tex]. We can cancel this common factor, provided [tex]\( x \neq 3 \)[/tex]:

[tex]\[ f(x) = \frac{x(x + 3)}{x + 1} \quad \text{for} \quad x \neq 3 \][/tex]

### Step 3: Evaluate the limit of the simplified expression

Now, substitute [tex]\( x = 3 \)[/tex] into the simplified expression:

[tex]\[ \lim_{x \to 3} \frac{x(x + 3)}{x + 1} = \frac{3(3 + 3)}{3 + 1} = \frac{3 \times 6}{4} = \frac{18}{4} = \frac{9}{2} \][/tex]

### Conclusion:

Thus, the limit of the given function as [tex]\( x \)[/tex] approaches 3 is:

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

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