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The given problem asks to determine the left-hand limit:

\[
\lim_{x \to 1^-} \left( x^2 - \frac{2}{x - 1} \right)
\]

### Step 1: Analyze the limit as \( x \to 1^- \)
As \( x \to 1^- \), notice that the term \( \frac{2}{x - 1} \) becomes problematic because \( x - 1 \) approaches 0 from the left (negative values), causing the expression to approach negative infinity:

\[
\frac{2}{x - 1} \to -\infty \quad \text{as} \quad x \to 1^-
\]

### Step 2: Examine the behavior of the other term
The term \( x^2 \) as \( x \to 1^- \) is straightforward:

\[
x^2 \to 1
\]

### Step 3: Combine the behaviors
Thus, the overall expression behaves as follows:

\[
x^2 - \frac{2}{x - 1} = 1 - (-\infty) = 1 + \infty = \infty
\]

### Conclusion:
The left-hand limit approaches positive infinity:

\[
\lim_{x \to 1^-} \left( x^2 - \frac{2}{x - 1} \right) = +\infty
\]

Would you like further clarification on any part of this solution, or a more detailed breakdown?

Here are 5 related questions for deeper understanding:
1. What is the difference between left-hand and right-hand limits?
2. How would you evaluate the limit if \( x \to 1^+ \)?
3. What are some common indeterminate forms encountered in limits?
4. Can limits always be computed algebraically, or are there cases where other techniques (like L'Hopital's Rule) are required?
5. How does the behavior of a rational function differ when approaching vertical asymptotes?

**Tip:** When dealing with limits that involve division by zero, always check if the function exhibits infinite behavior or if other indeterminate forms (like \( \frac{0}{0} \)) appear, which may require special techniques like factoring or L'Hopital's Rule.

Determine \( \lim_{x \to 1^-} \left( x^2 - \frac{2}{x - 1} \right) \).

Learn how to determine the left-hand limit of the expression \( x^2 - \frac{2}{x-1} \) as \( x \to 1^- \). This calculus problem explores the asymptotic behavior and infinite limits. Suitable for advanced high school students and college freshmen.

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