To find the limit of the expression $\frac{\frac{1}{4} + \frac{1}{x}}{4 + x}$ as $x \to -4$, let's proceed step by step.

Step 1: Write the expression

We are dealing with the limit: $\lim_{x \to -4} \frac{\frac{1}{4} + \frac{1}{x}}{4 + x}$

Step 2: Substitute $x = -4$ directly

Substituting $x = -4$ into the expression:

$\frac{\frac{1}{4} + \frac{1}{-4}}{4 - 4} = \frac{\frac{1}{4} - \frac{1}{4}}{0} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, so we need to simplify the expression or apply L'Hopital's Rule.

Step 3: Apply L'Hopital's Rule

Since we have an indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule. According to L'Hopital's Rule, we take the derivative of the numerator and denominator separately and then find the limit of the resulting expression.

Derivative of the numerator:

$\frac{d}{dx} \left( \frac{1}{4} + \frac{1}{x} \right) = 0 - \frac{1}{x^2} = -\frac{1}{x^2}$

Derivative of the denominator:

$\frac{d}{dx} \left( 4 + x \right) = 1$

Step 4: Rewrite the limit

Now we have the new limit: $\lim_{x \to -4} \frac{-\frac{1}{x^2}}{1}$ Substituting $x = -4$: $-\frac{1}{(-4)^2} = -\frac{1}{16}$

Final Answer:

The limit is: $\lim_{x \to -4} \frac{\frac{1}{4} + \frac{1}{x}}{4 + x} = -\frac{1}{16}$

Would you like more details on any step or further clarification?

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Here are 5 related questions to explore:

1. What are the conditions to apply L'Hopital's Rule?
2. How do you find limits that result in indeterminate forms other than $\frac{0}{0}$?
3. Can the limit change if we modify the numerator or denominator slightly?
4. How would this limit change if the denominator was $x - 4$ instead of $x + 4$?
5. What happens to the limit if $x \to \infty$ instead of $x \to -4$?

Tip: Always check if direct substitution works before using L'Hopital's Rule or other techniques.