Using the given function below, evaluate each of the of the following limits, if it exists.
Let g(x) = {

x if x < 0
x^2 if 0 < x ≤ 2
8 − x if x > 2

Evaluate each of the of the following limits, if it exists.

a. lim g(x)
x→ 0^+
b. lim g(x)
x→ 0^−
c. lim g(x)
x→ 0
d. lim g(x)
x→ 2^−
e. lim g(x)
x→ 2

Given the piecewise function \( g(x) \):

\[
g(x) = 
\begin{cases} 
x & \text{if } x < 0 \\
x^2 & \text{if } 0 < x \leq 2 \\
8 - x & \text{if } x > 2
\end{cases}
\]

We are tasked with evaluating various limits. Let’s go step by step.

### a. \(\lim_{x \to 0^+} g(x)\)

This represents the right-hand limit as \(x\) approaches 0 from values greater than 0. When \(x > 0\), the function definition is \(g(x) = x^2\). Therefore, we use the \(x^2\) expression to evaluate the limit:

\[
\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0
\]

### b. \(\lim_{x \to 0^-} g(x)\)

This represents the left-hand limit as \(x\) approaches 0 from values less than 0. When \(x < 0\), the function definition is \(g(x) = x\). Therefore, we use the \(x\) expression to evaluate the limit:

\[
\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} x = 0
\]

### c. \(\lim_{x \to 0} g(x)\)

The limit exists if and only if both the left-hand and right-hand limits are equal. From parts (a) and (b), we found that:

\[
\lim_{x \to 0^+} g(x) = 0 \quad \text{and} \quad \lim_{x \to 0^-} g(x) = 0
\]

Thus:

\[
\lim_{x \to 0} g(x) = 0
\]

### d. \(\lim_{x \to 2^-} g(x)\)

This represents the left-hand limit as \(x\) approaches 2 from values less than 2. When \(0 < x \leq 2\), the function definition is \(g(x) = x^2\). Therefore, we evaluate the limit using the \(x^2\) expression:

\[
\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4
\]

### e. \(\lim_{x \to 2} g(x)\)

The limit exists if both the left-hand and right-hand limits at \(x = 2\) are equal. We already found:

- \(\lim_{x \to 2^-} g(x) = 4\)

For the right-hand limit, when \(x > 2\), the function is \(g(x) = 8 - x\). Therefore:

\[
\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (8 - x) = 8 - 2 = 6
\]

Since the left-hand limit (\(4\)) does not equal the right-hand limit (\(6\)), the overall limit does **not exist** at \(x = 2\).

---

### Summary of Results:

a. \(\lim_{x \to 0^+} g(x) = 0\)

b. \(\lim_{x \to 0^-} g(x) = 0\)

c. \(\lim_{x \to 0} g(x) = 0\)

d. \(\lim_{x \to 2^-} g(x) = 4\)

e. \(\lim_{x \to 2} g(x)\) does not exist.

---

Would you like any more details or have any questions?

Here are 5 related questions for deeper exploration:
1. What is the value of \(\lim_{x \to 3} g(x)\)?
2. How does the function behave as \(x\) approaches negative infinity?
3. For what values of \(x\) is the function \(g(x)\) continuous?
4. Can you determine any points where \(g(x)\) is discontinuous?
5. How would the function \(g(x)\) change if the middle piece were defined as \(x^3\) instead of \(x^2\)?

**Tip**: Always check both left-hand and right-hand limits when determining if a limit exists at a given point.

Using the given function below, evaluate each of the following limits, if it exists.
Let g(x) = {

x if x < 0
x^2 if 0 < x ≤ 2
8 − x if x > 2

Evaluate each of the following limits, if it exists.
a. lim g(x) x→ 0^+
b. lim g(x) x→ 0^−
c. lim g(x) x→ 0
d. lim g(x) x→ 2^−
e. lim g(x) x→ 2

Explore how to evaluate the limits of a piecewise function defined by g(x) with different expressions for x < 0, 0 < x ≤ 2, and x > 2. This problem covers important calculus concepts, such as one-sided limits and continuity, with detailed solutions. Suitable for Grade 11-12 students.

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