To evaluate the limit

$\lim_{x \to -8} \frac{2x^2 + 15x - 8}{x^2 - 64}$

we follow these steps:

Step 1: Check for direct substitution

Substitute $x = -8$ directly into the function.

• The numerator is: $2(-8)^2 + 15(-8) - 8 = 2(64) - 120 - 8 = 128 - 120 - 8 = 0$

• The denominator is: $(-8)^2 - 64 = 64 - 64 = 0$

Both the numerator and denominator result in 0, so we have an indeterminate form $\frac{0}{0}$. This indicates we should simplify the expression by factoring.

Step 2: Factor the numerator and denominator

• The denominator $x^2 - 64$ is a difference of squares: $x^2 - 64 = (x - 8)(x + 8)$

• For the numerator $2x^2 + 15x - 8$, we'll factor it using the method of grouping or trial and error. After factoring: $2x^2 + 15x - 8 = (2x - 1)(x + 8)$

Step 3: Simplify the expression

Now we rewrite the original function with the factored terms: $\frac{(2x - 1)(x + 8)}{(x - 8)(x + 8)}$

Cancel the common factor $(x + 8)$: $\frac{2x - 1}{x - 8}$

Step 4: Apply the limit

Now, substitute $x = -8$ into the simplified expression: $\frac{2(-8) - 1}{-8 - 8} = \frac{-16 - 1}{-16} = \frac{-17}{-16} = \frac{17}{16}$

Thus, the limit is: $\lim_{x \to -8} \frac{2x^2 + 15x - 8}{x^2 - 64} = \frac{17}{16}$

Final Answer:

$\boxed{\frac{17}{16}}$

Would you like further explanation on any step?

5 Related Questions:

1. How do you handle limits involving indeterminate forms like $\frac{0}{0}$?
2. Can all quadratic expressions be factored? What are alternative methods if they can't?
3. What is the difference of squares, and how is it used in simplifying limits?
4. What other indeterminate forms are common in limits, and how are they resolved?
5. How can L'Hopital's Rule be applied to evaluate limits?

Tip:

Always check for factorable terms in both the numerator and denominator when you encounter indeterminate forms like $\frac{0}{0}$.