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]
