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The function \( h(x) \) is piecewise defined as follows:

\[
h(x) = 
\begin{cases} 
\frac{7x^2 + 3}{x}, & \text{if } x \leq 0, \\
\frac{1}{x} - 1, & \text{if } 0 < x \leq 2, \\
\frac{x^2 - 5}{2}, & \text{if } x > 2.
\end{cases}
\]

The problem asks to find the following limits:

1. \( \lim_{x \to 0^-} h(x) \)
2. \( \lim_{x \to 0^+} h(x) \)
3. \( \lim_{x \to 0} h(x) \)
4. \( \lim_{x \to 2^-} h(x) \)
5. \( \lim_{x \to 2^+} h(x) \)
6. \( \lim_{x \to 2} h(x) \)

Additionally, it requests to graph the function \( h(x) \), investigate if there are asymptotes, and finally analyze the continuity of the function.

Let's start by calculating the limits at \( x = 0 \) and \( x = 2 \).

### 1. Limit as \( x \to 0^- \) (from the left):
For \( x \leq 0 \), the function is \( h(x) = \frac{7x^2 + 3}{x} \).
To find the left-hand limit:
\[
\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} \frac{7x^2 + 3}{x}.
\]
Since the numerator approaches 3 and the denominator approaches 0 from the negative side, the limit tends to \( -\infty \).

### 2. Limit as \( x \to 0^+ \) (from the right):
For \( 0 < x \leq 2 \), the function is \( h(x) = \frac{1}{x} - 1 \).
To find the right-hand limit:
\[
\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \left( \frac{1}{x} - 1 \right).
\]
As \( x \to 0^+ \), \( \frac{1}{x} \to \infty \), so the limit is \( \infty \).

### 3. Limit as \( x \to 0 \):
Since \( \lim_{x \to 0^-} h(x) = -\infty \) and \( \lim_{x \to 0^+} h(x) = \infty \), the two one-sided limits do not agree. Therefore:
\[
\lim_{x \to 0} h(x) \text{ does not exist}.
\]

### 4. Limit as \( x \to 2^- \):
For \( 0 < x \leq 2 \), the function is \( h(x) = \frac{1}{x} - 1 \).
To find the left-hand limit:
\[
\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} \left( \frac{1}{x} - 1 \right) = \frac{1}{2} - 1 = -\frac{1}{2}.
\]

### 5. Limit as \( x \to 2^+ \):
For \( x > 2 \), the function is \( h(x) = \frac{x^2 - 5}{2} \).
To find the right-hand limit:
\[
\lim_{x \to 2^+} h(x) = \frac{2^2 - 5}{2} = \frac{4 - 5}{2} = -\frac{1}{2}.
\]

### 6. Limit as \( x \to 2 \):
Since \( \lim_{x \to 2^-} h(x) = -\frac{1}{2} \) and \( \lim_{x \to 2^+} h(x) = -\frac{1}{2} \), the two one-sided limits agree. Therefore:
\[
\lim_{x \to 2} h(x) = -\frac{1}{2}.
\]

### Continuity
- At \( x = 0 \), the function is **not continuous** because the limit does not exist.
- At \( x = 2 \), the function is **continuous** because the left-hand and right-hand limits are equal, and the function value at \( x = 2 \) is also \( h(2) = -\frac{1}{2} \).

### Asymptotes
The function has a vertical asymptote at \( x = 0 \), since the limits approach \( \infty \) and \( -\infty \) from different sides.

Would you like me to plot the graph of this function for further investigation or provide more details on specific parts?

Here are 5 related questions:
1. What are the key properties of piecewise functions in terms of continuity?
2. How do you determine if a function has a vertical asymptote?
3. What role does the limit at a point play in determining continuity?
4. How can we classify different types of discontinuities (e.g., jump, infinite)?
5. How can you verify if a limit at a point exists algebraically?

**Tip:** Always check both the left-hand and right-hand limits to fully understand the behavior of a piecewise function at specific points.

Hallar lim_{x→0-} h(x), lim_{x→0+} h(x), lim_{x→0} h(x), lim_{x→2-} h(x), lim_{x→2+} h(x), lim_{x→2} h(x). Hacer su gráfica e investigar si tiene asíntotas. Por último, ¿Qué puede decir de su continuidad?

Learn how to solve a problem involving a piecewise function with limits at x = 0 and x = 2. Analyze continuity and identify any asymptotes. A detailed step-by-step solution is provided, covering key calculus concepts like limits, continuity, and asymptotes. Suitable for undergraduate calculus students.

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