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The question is asking to evaluate the following limit:

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

To solve this limit from the left (\(x \to 1^-\)):

### Step-by-Step Solution:

1. **Substitute \(x = 1\):**

   Substituting \(x = 1\) directly into the function:

   \[
   \frac{1^2 + 1}{1 - 1} = \frac{2}{0}
   \]

   This gives a division by zero. Therefore, we need to check if the limit tends to \(+\infty\), \(-\infty\), or a finite value by approaching \(x = 1\) from the left-hand side.

2. **Behavior near \(x = 1^-\):**

   - As \(x \to 1\) from the left (\(x < 1\)), the denominator \(x - 1\) is negative because \(x < 1\).
   - The numerator \(x^2 + 1\) is always positive since \(x^2 + 1 > 0\) for all real \(x\).

   Therefore, for \(x \to 1^-\), we have:

   \[
   \frac{x^2 + 1}{x - 1} \to \frac{2}{-\text{small positive number}} \to -\infty
   \]

   The function tends towards \(-\infty\) as \(x\) approaches 1 from the left.

### Conclusion:
The limit is \(-\infty\), which corresponds to option **b**.

---

Would you like further clarification? Here are some related questions for practice:

1. What is the limit from the right (\(x \to 1^+\)) of the same function?
2. What happens to the function as \(x \to -\infty\)?
3. How do you apply L'Hopital's Rule to this kind of limit?
4. What is the difference between one-sided limits and two-sided limits?
5. How does the numerator \(x^2 + 1\) influence the behavior of the limit?

**Tip:** Always check if direct substitution leads to an indeterminate form like \(\frac{0}{0}\) or \(\frac{a}{0}\), as this often indicates the need for further analysis.

Nilai dari lim x→1− (x² + 1) / (x - 1) adalah...

Learn how to evaluate the left-hand limit of the rational function (x² + 1) / (x - 1) as x approaches 1. This problem involves key limit concepts, analyzing function behavior near undefined points, and applying one-sided limit theorems.

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