The given problem is:

$\lim_{x \to 3} \frac{x^2 + 4x}{x^2 - 2x - 3}$

Step 1: Factor the denominator

First, let's factor the denominator $x^2 - 2x - 3$: $x^2 - 2x - 3 = (x - 3)(x + 1)$

Step 2: Check for direct substitution

If we substitute $x = 3$ directly into the expression:

• Numerator: $x^2 + 4x = 9 + 12 = 21$
• Denominator: $(3^2) - 2(3) - 3 = 9 - 6 - 3 = 0$

This results in a $\frac{21}{0}$, which means direct substitution gives an undefined result. However, this doesn't mean the limit does not exist; we need to explore further.

Step 3: Simplifying the expression

The expression does not simplify easily, but since substituting $x = 3$ yields a non-zero numerator and a zero denominator, it suggests that this limit is infinite. Specifically, we can conclude that:

$\lim_{x \to 3} \frac{x^2 + 4x}{x^2 - 2x - 3} = \infty$

Final Answer:

The limit does not exist in the traditional sense but tends toward infinity.

Would you like more details on this? Here are 5 related questions for further practice:

1. What is the limit if the denominator does not factor nicely?
2. How do you handle limits that tend toward zero from the left and right sides?
3. Can you find the limit when $x$ tends to 3 for a different rational function?
4. What are the rules for dealing with indeterminate forms like $\frac{0}{0}$?
5. How can L'Hopital's Rule help with similar limits?

Tip: Always check if the function factors or simplifies before applying limit laws.