To solve the limit $\lim_{{x \to -4}} \left( \frac{\frac{1}{4} - \frac{1}{x}}{4 + x} \right)$, we'll follow these steps:

1. Simplify the expression inside the limit:

We start by simplifying the numerator:

$\frac{1}{4} - \frac{1}{x} = \frac{x - 4}{4x}$

Thus, the expression becomes:

$\frac{\frac{x - 4}{4x}}{4 + x} = \frac{x - 4}{4x(4 + x)}$

2. Apply the limit:

Now, substitute $x = -4$ directly into the expression:

$\lim_{{x \to -4}} \frac{x - 4}{4x(4 + x)} = \frac{-4 - 4}{4(-4)(4 - 4)}$

This simplifies to:

$\frac{-8}{-64} = \frac{1}{8}$

However, notice that substituting $x = -4$ into the denominator $4 + x$ results in zero. This suggests that direct substitution is not possible and we need to use L'Hôpital's Rule.

3. Use L'Hôpital's Rule:

L'Hôpital's Rule applies because both the numerator and denominator approach 0 as $x \to -4$. We differentiate the numerator and the denominator separately:

• Differentiate the numerator: $\text{Numerator: } \frac{d}{dx}\left(\frac{1}{4} - \frac{1}{x}\right) = \frac{d}{dx}\left(-\frac{1}{x}\right) = \frac{1}{x^2}$

• Differentiate the denominator: $\text{Denominator: } \frac{d}{dx}(4 + x) = 1$

The expression simplifies to:

$\lim_{{x \to -4}} \frac{1/x^2}{1} = \frac{1}{(-4)^2} = \frac{1}{16}$

So, the final answer is:

$\boxed{\frac{1}{16}}$

Would you like more details on any part of this solution or have any questions?

Here are some related questions:

1. How do you determine when to use L'Hôpital's Rule?
2. What are the conditions for applying L'Hôpital's Rule?
3. How does differentiating the numerator and denominator separately work in L'Hôpital's Rule?
4. Why is direct substitution not always possible in limits?
5. What happens if L'Hôpital's Rule gives another indeterminate form after the first application?

Tip: Always check if direct substitution in a limit results in an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$) to decide if L'Hôpital's Rule is applicable.