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The problem you provided is:

\[
\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{1}{x^3 - 2x^2} \right]
\]

### Step-by-Step Solution:

1. **Evaluate each part as \( x \to 2 \):**
   - The first term, \( \frac{1}{x - 2} \), tends to infinity as \( x \) approaches 2.
   - The second term, \( \frac{1}{x^3 - 2x^2} \), also tends to infinity as \( x \to 2 \), since at \( x = 2 \), the denominator becomes \( 8 - 8 = 0 \).

2. **Simplifying the denominator of the second fraction:**
   The second term can be factored:
   \[
   x^3 - 2x^2 = x^2(x - 2)
   \]
   So the second term becomes:
   \[
   \frac{1}{x^2(x - 2)}
   \]

3. **Rewrite the limit:**
   \[
   \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{1}{x^2(x - 2)} \right]
   \]
   Factor out \( \frac{1}{x - 2} \):
   \[
   \lim_{x \to 2} \frac{1}{x - 2} \left[ 1 - \frac{1}{x^2} \right]
   \]

4. **Simplify the expression inside the brackets:**
   \[
   1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} = \frac{(x - 1)(x + 1)}{x^2}
   \]

5. **Final limit:**
   Now the expression becomes:
   \[
   \lim_{x \to 2} \frac{(x - 1)(x + 1)}{x^2(x - 2)}
   \]
   As \( x \to 2 \), substitute the value:
   \[
   = \frac{(2 - 1)(2 + 1)}{2^2(2 - 2)} = \frac{3}{4 \times 0} = \infty
   \]
   Thus, the limit does not exist because it tends to infinity.

### Conclusion:
The limit tends to infinity, meaning it does not exist in a finite sense.

Do you want more details or have any questions? Here are five related questions to expand on this topic:

1. How do you apply L'Hopital's Rule to solve indeterminate limits?
2. What are the common factorizations used when simplifying limits?
3. How can you recognize if a limit approaches infinity or a finite value?
4. Why is factoring useful in solving limits involving polynomials?
5. What strategies can you use to approach limits that involve subtraction of terms?

**Tip:** Always check if the limit results in an indeterminate form like \( \frac{0}{0} \) or \( \infty - \infty \), as these often require more advanced techniques like factoring or applying L'Hopital's Rule.

Evaluate the limit: \( \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{1}{x^3 - 2x^2} \right] \)

Learn how to evaluate the limit involving a subtraction of terms: \( \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{1}{x^3 - 2x^2} \right] \). This step-by-step solution covers key mathematical concepts like factoring and infinity in limits, suitable for Grades 10-12.

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