Construct a table and find the indicated limit.
If ​h(x)equalsStartFraction StartRoot x EndRoot plus 2 Over x minus 8 EndFraction
​, then find ModifyingBelow lim With x right arrow 8 h left parenthesis x right parenthesis.
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Part 1
Complete the table below.
x
7.9
7.99
7.999
8.001
8.01
8.1
h left parenthesis x right parenthesis
  
enter your response here
  
enter your response here
  
enter your response here
  
enter your response here
  
enter your response here
  
enter your response here
​(Type integers or decimals rounded to four decimal places as​ needed.)

We are given the function \( h(x) = \frac{\sqrt{x} + 2}{x - 8} \), and the goal is to estimate the limit \( \lim_{x \to 8} h(x) \) by evaluating the function at points around \( x = 8 \).

Let's evaluate \( h(x) \) at the given points. We will use the following values for \( x \): 7.9, 7.99, 7.999, 8.001, 8.01, and 8.1.

For each \( x \), we will substitute the value into the function \( h(x) = \frac{\sqrt{x} + 2}{x - 8} \) and calculate it to four decimal places.

### Calculations:

1. **For \( x = 7.9 \):**
   \[
   h(7.9) = \frac{\sqrt{7.9} + 2}{7.9 - 8} = \frac{\sqrt{7.9} + 2}{-0.1}
   \]
   Using \( \sqrt{7.9} \approx 2.8107 \):
   \[
   h(7.9) = \frac{2.8107 + 2}{-0.1} = \frac{4.8107}{-0.1} = -48.107
   \]

2. **For \( x = 7.99 \):**
   \[
   h(7.99) = \frac{\sqrt{7.99} + 2}{7.99 - 8} = \frac{\sqrt{7.99} + 2}{-0.01}
   \]
   Using \( \sqrt{7.99} \approx 2.8270 \):
   \[
   h(7.99) = \frac{2.8270 + 2}{-0.01} = \frac{4.8270}{-0.01} = -482.700
   \]

3. **For \( x = 7.999 \):**
   \[
   h(7.999) = \frac{\sqrt{7.999} + 2}{7.999 - 8} = \frac{\sqrt{7.999} + 2}{-0.001}
   \]
   Using \( \sqrt{7.999} \approx 2.8284 \):
   \[
   h(7.999) = \frac{2.8284 + 2}{-0.001} = \frac{4.8284}{-0.001} = -4828.400
   \]

4. **For \( x = 8.001 \):**
   \[
   h(8.001) = \frac{\sqrt{8.001} + 2}{8.001 - 8} = \frac{\sqrt{8.001} + 2}{0.001}
   \]
   Using \( \sqrt{8.001} \approx 2.8284 \):
   \[
   h(8.001) = \frac{2.8284 + 2}{0.001} = \frac{4.8284}{0.001} = 4828.400
   \]

5. **For \( x = 8.01 \):**
   \[
   h(8.01) = \frac{\sqrt{8.01} + 2}{8.01 - 8} = \frac{\sqrt{8.01} + 2}{0.01}
   \]
   Using \( \sqrt{8.01} \approx 2.8297 \):
   \[
   h(8.01) = \frac{2.8297 + 2}{0.01} = \frac{4.8297}{0.01} = 482.970
   \]

6. **For \( x = 8.1 \):**
   \[
   h(8.1) = \frac{\sqrt{8.1} + 2}{8.1 - 8} = \frac{\sqrt{8.1} + 2}{0.1}
   \]
   Using \( \sqrt{8.1} \approx 2.8460 \):
   \[
   h(8.1) = \frac{2.8460 + 2}{0.1} = \frac{4.8460}{0.1} = 48.460
   \]

### Table of values:
\[
\begin{array}{|c|c|}
\hline
x & h(x) \\
\hline
7.9 & -48.1070 \\
7.99 & -482.7000 \\
7.999 & -4828.4000 \\
8.001 & 4828.4000 \\
8.01 & 482.9700 \\
8.1 & 48.4600 \\
\hline
\end{array}
\]

### Conclusion:
The function \( h(x) \) approaches \( \pm \infty \) as \( x \to 8 \), indicating a vertical asymptote at \( x = 8 \). Therefore, the limit does not exist in a finite sense.

---

Let me know if you would like more details or explanations.

Here are 5 related questions you could explore:
1. What does it mean for a function to have a vertical asymptote?
2. How do we interpret limits that approach infinity?
3. What techniques can we use to evaluate limits analytically instead of using tables?
4. Can we find limits at infinity for other types of functions?
5. What is the difference between one-sided and two-sided limits?

**Tip:** When a function's denominator approaches zero, always check the behavior of the numerator to identify possible vertical asymptotes.

Construct a table and find the indicated limit. If h(x) equals sqrt(x) + 2 over x - 8, then find lim as x approaches 8 h(x). Complete the table below.
x: 7.9, 7.99, 7.999, 8.001, 8.01, 8.1
h(x): [response needed]
(Type integers or decimals rounded to four decimal places as needed.)

Learn how to evaluate the limit of h(x) = (sqrt(x) + 2) / (x - 8) as x approaches 8. The solution involves constructing a table with values close to x = 8 and determining whether the limit exists. This problem covers key concepts such as limits, vertical asymptotes, and function behavior. Suitable for Grades 10-12.

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