The given limit is:

$\lim_{x \to 4} \frac{-1}{4 - x}$

Solution:

As $x$ approaches 4 from either side, the denominator $(4 - x)$ approaches 0. Let's analyze the behavior from both sides:

1. As $x \to 4^+$ (from the right):

   • When $x$ is slightly greater than 4, $4 - x$ is a small negative number.
   • Therefore, $\frac{-1}{4 - x}$ becomes a large positive number because you are dividing a negative number by a small negative value.

2. As $x \to 4^-$ (from the left):

   • When $x$ is slightly less than 4, $4 - x$ is a small positive number.
   • Therefore, $\frac{-1}{4 - x}$ becomes a large negative number because you are dividing a negative number by a small positive value.

Conclusion:

• As $x \to 4^+$, the limit tends to $+\infty$.
• As $x \to 4^-$, the limit tends to $-\infty$.

Thus, the limit does not exist because the left-hand and right-hand limits approach different infinities.

However, if you're considering the behavior from one specific side:

• If you meant $x \to 4^+$ (from the right), the limit is $+\infty$.
• If you meant $x \to 4^-$ (from the left), the limit is $-\infty$.

Would you like more details or explanations on any part of this?

Here are 5 related questions to expand:

1. What is the formal definition of a limit at infinity?
2. How do we determine left-hand and right-hand limits?
3. What happens if both left-hand and right-hand limits go to the same infinity?
4. How does L'Hôpital's Rule apply to limits with indeterminate forms?
5. Can limits at infinity be finite?

Tip: When evaluating limits, always check for different behaviors on either side of the point, especially when the denominator tends to zero.