We want to compute the limit of \( \frac{e^{-x}}{x-2} \) as \( x \to 2 \).

The expression is:

\[
\lim_{x \to 2} \frac{e^{-x}}{x - 2}
\]

### Substituting \( x = 2 \) into the function:
If we directly substitute \( x = 2 \), we get:

\[
\frac{e^{-2}}{2 - 2} = \frac{e^{-2}}{0}
\]

This results in a division by zero, which means we need to further investigate the limit. Since this involves a division by zero, we'll explore the behavior of the limit from both sides of \( x = 2 \).

### Case 1: \( x \to 2^+ \) (approaching 2 from the right)
As \( x \to 2^+ \), the denominator \( x - 2 \) becomes a small positive number, while \( e^{-x} \) is positive and finite. So, the expression tends towards positive infinity.

### Case 2: \( x \to 2^- \) (approaching 2 from the left)
As \( x \to 2^- \), the denominator \( x - 2 \) becomes a small negative number, while \( e^{-x} \) is still positive and finite. Thus, the expression tends towards negative infinity.

### Conclusion:
The limit does not exist because it tends to \( +\infty \) as \( x \to 2^+ \) and \( -\infty \) as \( x \to 2^- \). Therefore:

\[
\lim_{x \to 2} \frac{e^{-x}}{x - 2} = \text{does not exist}.
\]

Would you like more details or clarification?

Here are 5 related questions:
1. What are limits from the left and right?
2. How do you handle indeterminate forms like \( \frac{0}{0} \)?
3. Can L'Hopital's Rule apply to this type of problem?
4. What is the behavior of exponential functions as \( x \to \infty \)?
5. How do limits help in understanding asymptotic behavior?

**Tip:** When evaluating limits involving division by zero, check the behavior from both sides to determine if the limit exists.

Explore the limit of the function (e^-x)/(x-2) as x approaches 2. This problem involves analyzing one-sided limits and understanding the behavior of functions near points of discontinuity. Suitable for students in Grades 11-12 studying pre-calculus or calculus.

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