(4) \( \lim _{x \rightarrow-4} \frac{\frac{1}{4}+\frac{1}{x}}{4+x} \)
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To compute the limit \( \lim _{x \rightarrow-4} \frac{\frac{1}{4}+\frac{1}{x}}{4+x} \), we start by substituting \( x = -4 \): Substituting \( x = -4 \) into the denominator gives: \[ 4 + (-4) = 0 \] This indicates the expression is undefined at \( x = -4 \). Next, we check the numerator: \[ \frac{1}{4} + \frac{1}{-4} = \frac{1}{4} - \frac{1}{4} = 0 \] Thus, we have a form \( \frac{0}{0} \), which suggests that we can simplify the function. Let's simplify \( \frac{\frac{1}{4} + \frac{1}{x}}{4+x} \): The numerator can be rewritten as: \[ \frac{1}{4} + \frac{1}{x} = \frac{x + 4}{4x} \] So the original expression becomes: \[ \frac{\frac{x + 4}{4x}}{4 + x} \] Now we can simplify this further: \[ = \frac{x + 4}{4x(4+x)} \] Now cancel \( x + 4 \) from the numerator and denominator: \[ = \frac{1}{4x} \text{ for } x \neq -4 \] Next, we calculate the limit: \[ \lim_{x \to -4} \frac{1}{4x} = \frac{1}{4 \cdot (-4)} = \frac{1}{-16} = -\frac{1}{16} \] Thus, the final answer is: \[ \boxed{-\frac{1}{16}} \]