The given problem is:

$\lim_{x \to -4} \frac{\sqrt{x + 12} - \sqrt{4 - x}}{x^2 + 2x - 8}$

Let's first evaluate this limit step-by-step.

Step 1: Check for the indeterminate form

Substituting $x = -4$ directly into the expression:

$\sqrt{-4 + 12} - \sqrt{4 - (-4)} = \sqrt{8} - \sqrt{8} = 0$

And in the denominator: $(-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0$

Hence, we have the indeterminate form $\frac{0}{0}$. We will need to simplify the expression further.

Step 2: Rationalizing the numerator

To handle the indeterminate form, we rationalize the numerator: $\frac{\sqrt{x + 12} - \sqrt{4 - x}}{x^2 + 2x - 8} \times \frac{\sqrt{x + 12} + \sqrt{4 - x}}{\sqrt{x + 12} + \sqrt{4 - x}}$ This gives us: $= \frac{(\sqrt{x + 12})^2 - (\sqrt{4 - x})^2}{(x^2 + 2x - 8)(\sqrt{x + 12} + \sqrt{4 - x})}$ The numerator simplifies to: $(x + 12) - (4 - x) = x + 12 - 4 + x = 2x + 8$ Thus, the expression becomes: $\frac{2(x + 4)}{(x^2 + 2x - 8)(\sqrt{x + 12} + \sqrt{4 - x})}$

Step 3: Factor the denominator

The denominator $x^2 + 2x - 8$ factors as: $(x + 4)(x - 2)$ Therefore, the expression is: $\frac{2(x + 4)}{(x + 4)(x - 2)(\sqrt{x + 12} + \sqrt{4 - x})}$ We can cancel out the $(x + 4)$ term (since $x \to -4$): $= \frac{2}{(x - 2)(\sqrt{x + 12} + \sqrt{4 - x})}$

Step 4: Substituting $x = -4$

Now, substituting $x = -4$: $= \frac{2}{(-4 - 2)(\sqrt{-4 + 12} + \sqrt{4 - (-4)})}$ $= \frac{2}{(-6)(\sqrt{8} + \sqrt{8})}$ $= \frac{2}{-6 \cdot 2\sqrt{8}}$ $= \frac{2}{-12\sqrt{8}}$ $= -\frac{1}{6\sqrt{8}}$ Simplify $\sqrt{8} = 2\sqrt{2}$: $= -\frac{1}{6 \cdot 2\sqrt{2}} = -\frac{1}{12\sqrt{2}}$ Further simplifying, $= -\frac{\sqrt{2}}{24}$

Final Answer

$\lim_{x \to -4} \frac{\sqrt{x + 12} - \sqrt{4 - x}}{x^2 + 2x - 8} = -\frac{\sqrt{2}}{24}$

Would you like more details or have any questions?

Related Questions

1. How do you handle indeterminate forms in limits?
2. What are other ways to rationalize complex expressions in limits?
3. How do you factor polynomials like $x^2 + 2x - 8$?
4. What are the advantages of rationalizing the numerator vs. the denominator in limit problems?
5. How would this limit change if $x \to 4$ instead?

Tip

Always check for factorization opportunities in indeterminate limit forms—they often lead to cancellations that simplify the expression.