To solve the limit problem, let’s analyze it step-by-step. The expression is:

$\lim_{{x \to 12}} \frac{x^2 + 3x - 180}{x^2 - 144}$

Step 1: Substitute $x = 12$

By direct substitution, we get:

$\frac{12^2 + 3 \cdot 12 - 180}{12^2 - 144} = \frac{144 + 36 - 180}{144 - 144} = \frac{0}{0}$

Since this yields an indeterminate form ($\frac{0}{0}$), we need to simplify the expression further.

Step 2: Factor the Numerator and Denominator

1. Numerator: $x^2 + 3x - 180$

   • To factor this, we look for two numbers that multiply to $-180$ and add up to $3$.
   • The factors are $15$ and $-12$, so we can write: $x^2 + 3x - 180 = (x + 15)(x - 12)$

2. Denominator: $x^2 - 144$

   • This is a difference of squares: $x^2 - 144 = (x + 12)(x - 12)$

Step 3: Simplify the Expression

The limit now becomes:

$\lim_{{x \to 12}} \frac{(x + 15)(x - 12)}{(x + 12)(x - 12)}$

We can cancel out the $(x - 12)$ terms from the numerator and the denominator, assuming $x \neq 12$:

$\lim_{{x \to 12}} \frac{x + 15}{x + 12}$

Step 4: Substitute $x = 12$ Again

Now that the expression is simplified, we substitute $x = 12$:

$\frac{12 + 15}{12 + 12} = \frac{27}{24} = \frac{9}{8}$

Conclusion

The limit is:

$\lim_{{x \to 12}} \frac{x^2 + 3x - 180}{x^2 - 144} = \frac{9}{8}$

Would you like further details on any part of the solution or have any other questions?

Here are five related questions for further exploration:

1. How can we apply L'Hôpital's Rule to solve indeterminate forms?
2. What are other techniques for simplifying limits involving polynomials?
3. How can limits help us understand continuity at a given point?
4. How does the concept of the difference of squares help in simplifying expressions?
5. What are some common factorization techniques useful in limit problems?

Tip: When you encounter a $\frac{0}{0}$ indeterminate form, consider factoring or simplifying the expression before trying other methods like L'Hôpital's Rule.