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The image contains a mathematical problem asking to find the limit of a piecewise function \( f(x) \) as \( x \to 2 \).

The piecewise function \( f(x) \) is defined as:

\[
f(x) = 
\begin{cases} 
x^3 - 2 & \text{if } x < 2 \\
4x + 2 & \text{if } x \geq 2
\end{cases}
\]

The task is to determine if the limit of \( f(x) \) exists as \( x \to 2 \).

### Step-by-Step Solution:

To determine if the limit exists, we need to find the left-hand limit (LHL) and right-hand limit (RHL) of \( f(x) \) as \( x \to 2 \).

1. **Left-hand limit (LHL) as \( x \to 2 \):**
   \[
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x^3 - 2)
   \]

   Substituting \( x = 2 \):
   \[
   = (2)^3 - 2 = 8 - 2 = 6
   \]

2. **Right-hand limit (RHL) as \( x \to 2 \):**
   \[
   \lim_{x \to 2^+} f(x) = \lim_{x \to 2} (4x + 2)
   \]

   Substituting \( x = 2 \):
   \[
   = 4(2) + 2 = 8 + 2 = 10
   \]

3. **Compare LHL and RHL:**

   Since the left-hand limit (6) is not equal to the right-hand limit (10), the limit does not exist:

\[
\lim_{x \to 2} f(x) \text{ does not exist.}
\]

### Conclusion:
The limit of the function \( f(x) \) as \( x \to 2 \) does not exist because the left-hand limit and the right-hand limit are not equal.

Would you like more details or have any questions? 

Here are 5 related questions to explore further:
1. What is a piecewise function, and how do we define it?
2. How do we calculate left-hand and right-hand limits in general?
3. Can we determine the continuity of a piecewise function at a given point?
4. How do we find the derivative of a piecewise function?
5. What is the significance of limits in calculus and real-world applications?

**Tip:** Remember that for a limit to exist at a point, both the left-hand and right-hand limits must be equal.

Find the limit of the piecewise function f(x) as x approaches 2, where f(x) is defined as:
f(x) = 
  { x^3 - 2 if x < 2
  { 4x + 2 if x >= 2

Learn how to find the limit of a piecewise function as x approaches 2, involving both left-hand and right-hand limits. Detailed step-by-step solution provided. Suitable for high school calculus students.

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