limit as x goes to 2 from the left (x^3|x-2|)/(x-2)

We are tasked with finding the limit:

\[
\lim_{{x \to 2^-}} \frac{x^3 |x - 2|}{x - 2}
\]

### Step 1: Understanding the absolute value function
Since \(x \to 2^-\) (from the left), we know that \(x < 2\). Therefore:

\[
|x - 2| = -(x - 2) \quad \text{for} \quad x < 2
\]

So the expression becomes:

\[
\lim_{{x \to 2^-}} \frac{x^3 (-(x - 2))}{x - 2}
\]

### Step 2: Simplifying the expression
We can factor out \((x - 2)\) from the numerator and denominator:

\[
\lim_{{x \to 2^-}} \frac{-x^3(x - 2)}{x - 2}
\]

Canceling \((x - 2)\) in the numerator and denominator, we are left with:

\[
\lim_{{x \to 2^-}} -x^3
\]

### Step 3: Evaluating the limit
Now, simply substitute \(x = 2\):

\[
-2^3 = -8
\]

Thus, the limit is:

\[
\boxed{-8}
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. What happens to the limit if \(x\) approaches 2 from the right?
2. How do we deal with limits involving absolute value functions in general?
3. What is the significance of canceling terms in limit problems?
4. How does the behavior of limits change for different types of discontinuities?
5. How can we compute limits using L'Hopital's Rule?

**Tip:** When dealing with absolute values in limits, always carefully consider the direction of the approach to properly handle the sign.

Explore the step-by-step solution for finding the limit of (x^3|x-2|)/(x-2) as x approaches 2 from the left. This problem deals with limits, absolute value, and polynomial functions, suitable for students in Grades 11-12 or college-level calculus.

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