To solve the given limit, let's first clarify the expression. Based on your description, I assume you're asking for:

$\lim_{x \to 16} \frac{4 - x^2}{16x - x^2}$

Steps to Solve:

1. Factor the numerator and denominator: The numerator $4 - x^2$ is a difference of squares: $4 - x^2 = (2 - x)(2 + x)$

   The denominator $16x - x^2$ can be factored as: $16x - x^2 = x(16 - x)$

2. Rewrite the expression: Substituting the factored forms into the original limit expression: $\frac{(2 - x)(2 + x)}{x(16 - x)}$

3. Evaluate the limit: Now, substitute $x = 16$ directly: $\frac{(2 - 16)(2 + 16)}{16(16 - 16)} = \frac{(-14)(18)}{16 \times 0} = \frac{-252}{0}$ The denominator becomes zero while the numerator is non-zero, leading to the conclusion that the limit does not exist because it approaches infinity (or negative infinity, depending on the direction of approach).

Conclusion:

The limit does not exist (it tends to negative infinity).

Would you like further clarification on this, or do you have any additional questions?

Related Questions:

1. What happens when both numerator and denominator tend to zero in limits?
2. How do we handle limits that lead to an indeterminate form?
3. Can you explain L'Hopital's Rule for limits involving indeterminate forms?
4. What other types of limits result in non-existence?
5. How do conjugates help simplify limits in complex fractions?

Tip:

Always factor expressions before substituting in limits to avoid missing potential simplifications.