Let f(x) = 
5x + 1
x2 + 7x − 18
.

Determine each of the following values. If an answer does not exist, enter DNE.

lim 
x→3− 
f(x) = 
4/3
 
Correct: Your answer is correct.


lim 
x→3+ 
f(x) = 
4/3
 
Correct: Your answer is correct.


lim 
x→2− 
f(x) = 
-INFINITY
 
Incorrect: Your answer is incorrect.


lim 
x→2+ 
f(x) = 
INFINITY
 
Incorrect: Your answer is incorrect.

To determine the limits of the function \( f(x) = \frac{5x + 1}{x^2 + 7x - 18} \) as \( x \) approaches certain values, let's analyze the behavior of the function around those points.

First, let's factor the denominator to identify potential issues or points where the function might be undefined:

\[ x^2 + 7x - 18 = (x + 9)(x - 2) \]

This shows that the function has vertical asymptotes at \( x = -9 \) and \( x = 2 \). These are the points where the denominator is zero and the function may go to \( \pm \infty \).

### Given Answers:
1. \(\lim_{x \to 3^-} f(x) = \frac{4}{3}\)
2. \(\lim_{x \to 3^+} f(x) = \frac{4}{3}\)

Both these answers are correct.

### Checking Remaining Limits:
3. \(\lim_{x \to 2^-} f(x)\)
4. \(\lim_{x \to 2^+} f(x)\)

Since \( x = 2 \) is a vertical asymptote, let's determine the behavior of the function as \( x \) approaches 2 from the left and from the right.

#### Limit as \( x \to 2^- \):
As \( x \) approaches 2 from the left (\( x \to 2^- \)):
\[ f(x) = \frac{5x + 1}{(x + 9)(x - 2)} \]

When \( x \) is slightly less than 2, the term \( (x - 2) \) is negative and very small, causing the fraction to become very large in magnitude but negative in sign:
\[ \lim_{x \to 2^-} \frac{5x + 1}{(x + 9)(x - 2)} = -\infty \]

So, the correct answer is \( -\infty \).

#### Limit as \( x \to 2^+ \):
As \( x \) approaches 2 from the right (\( x \to 2^+ \)):
\[ f(x) = \frac{5x + 1}{(x + 9)(x - 2)} \]

When \( x \) is slightly greater than 2, the term \( (x - 2) \) is positive and very small, causing the fraction to become very large in magnitude but positive in sign:
\[ \lim_{x \to 2^+} \frac{5x + 1}{(x + 9)(x - 2)} = \infty \]

So, the correct answer is \( \infty \).

### Summary of Correct Answers:
1. \(\lim_{x \to 3^-} f(x) = \frac{4}{3}\)
2. \(\lim_{x \to 3^+} f(x) = \frac{4}{3}\)
3. \(\lim_{x \to 2^-} f(x) = -\infty\)
4. \(\lim_{x \to 2^+} f(x) = \infty\)

Would you like more details on any of these steps or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you determine the horizontal asymptotes of a rational function?
2. What is the significance of the numerator in the limit calculation for rational functions?
3. How do you find the oblique asymptotes of a function?
4. How can you use limits to find the points of discontinuity in a function?
5. What is the difference between removable and non-removable discontinuities?

**Tip:** When evaluating limits approaching a vertical asymptote, always check the signs of the terms involved to determine if the limit goes to \( \infty \) or \( -\infty \).

Let f(x) = 5x + 1 / (x^2 + 7x − 18). Determine each of the following values. If an answer does not exist, enter DNE. lim x→3− f(x) = 4/3, lim x→3+ f(x) = 4/3, lim x→2− f(x) = -INFINITY, lim x→2+ f(x) = INFINITY.

Learn how to evaluate the limits of the rational function f(x) = (5x + 1) / (x^2 + 7x − 18) as x approaches 3 and 2. This problem involves key concepts such as limits, rational functions, and vertical asymptotes. Detailed step-by-step solutions included, suitable for students in Grades 11-12.

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