The limit expression provided is:

$\lim_{{x \to 2^+}} \frac{1 - x}{x - 2}$

We need to evaluate this limit as $x$ approaches 2 from the right ($x \to 2^+$).

Step-by-Step Solution:

1. Rewriting the Expression: The denominator $x - 2$ approaches 0 as $x \to 2$. Specifically, since we are approaching from the right ($x > 2$), $x - 2$ is slightly positive.

   The numerator $1 - x$ approaches $1 - 2 = -1$ as $x \to 2$.

2. Behavior Near $x = 2$:

   • Numerator: $1 - x \to -1$
   • Denominator: $x - 2 \to 0^+$ (slightly positive)

3. Evaluating the Limit: The expression simplifies to:

   $\lim_{{x \to 2^+}} \frac{-1}{x - 2}$

   As $x$ approaches 2 from the right, $x - 2$ approaches a very small positive number, making the fraction $\frac{-1}{x - 2}$ approach negative infinity:

   $\lim_{{x \to 2^+}} \frac{-1}{x - 2} = -\infty$

Conclusion:

The limit is:

$\lim_{{x \to 2^+}} \frac{1 - x}{x - 2} = -\infty$

Would you like more details on any step? Here are some related questions:

1. What happens if we take the limit as $x \to 2^-$ (from the left)?
2. How do you evaluate limits involving infinity?
3. Can this limit be solved using L'Hopital's Rule?
4. What is the significance of approaching a point from the left or the right in limits?
5. How can we apply this concept to piecewise functions?

Tip: Always consider the direction of approach when evaluating limits, especially when a term goes to zero.