The problem asks to compute the limit:

$\lim_{x \to 2} \frac{4 - x^2}{3 - \sqrt{x^2 + 5}}$

Step 1: Direct Substitution

Let's try to substitute $x = 2$ directly into the expression:

• Numerator: $4 - (2)^2 = 4 - 4 = 0$
• Denominator: $3 - \sqrt{(2)^2 + 5} = 3 - \sqrt{4 + 5} = 3 - \sqrt{9} = 3 - 3 = 0$

Substituting gives the indeterminate form $\frac{0}{0}$. Therefore, we must simplify the expression further by algebraic manipulation or L'Hopital's Rule.

Step 2: Simplify the Expression

The numerator can be factored: $4 - x^2 = (2 - x)(2 + x)$ The expression becomes: $\frac{(2 - x)(2 + x)}{3 - \sqrt{x^2 + 5}}$

Step 3: Use L'Hopital's Rule

Since direct substitution leads to an indeterminate form, we can apply L'Hopital's Rule. This requires us to take the derivative of the numerator and denominator.

Derivative of the Numerator:

$\frac{d}{dx} (4 - x^2) = -2x$

Derivative of the Denominator:

$\frac{d}{dx} \left( 3 - \sqrt{x^2 + 5} \right) = - \frac{x}{\sqrt{x^2 + 5}}$

Step 4: Apply L'Hopital's Rule

Now applying L'Hopital's Rule, we compute the new limit:

$\lim_{x \to 2} \frac{-2x}{-\frac{x}{\sqrt{x^2 + 5}}}$

Simplifying the expression:

$\lim_{x \to 2} \frac{-2x \cdot \sqrt{x^2 + 5}}{-x} = \lim_{x \to 2} 2 \sqrt{x^2 + 5}$

Now substitute $x = 2$:

$2 \sqrt{2^2 + 5} = 2 \sqrt{4 + 5} = 2 \times 3 = 6$

Final Answer:

The value of the limit is $\boxed{6}$.

Let me know if you need more details, or have any questions. Here are some related questions to explore:

1. How does L'Hopital's Rule help in solving indeterminate forms?
2. Can you always apply L'Hopital's Rule, or are there restrictions?
3. What are some other methods for resolving $\frac{0}{0}$ indeterminate forms?
4. How can you factor $a^2 - b^2$ to simplify expressions?
5. When does substituting values directly in limits give valid results?

Tip: Whenever you encounter the $\frac{0}{0}$ form in limits, check if the expression can be factored or if L'Hopital's Rule applies.