By using the theory of limits solve this :y=x+1/2x-3

To solve the problem \( y = \frac{x + 1}{2x - 3} \) using the theory of limits, we need to explore the behavior of the function as \( x \) approaches certain key values. In particular, let's analyze the limits as \( x \) approaches infinity, negative infinity, and the point where the denominator becomes zero.

### 1. Limit as \( x \to \infty \):
We are interested in understanding the behavior of the function when \( x \) becomes very large (positive infinity). So we compute:

\[
\lim_{x \to \infty} \frac{x + 1}{2x - 3}
\]

To simplify, divide both the numerator and the denominator by \( x \):

\[
\lim_{x \to \infty} \frac{\frac{x + 1}{x}}{\frac{2x - 3}{x}} = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{2 - \frac{3}{x}}
\]

As \( x \to \infty \), the terms \( \frac{1}{x} \) and \( \frac{3}{x} \) approach zero, so the limit simplifies to:

\[
\frac{1 + 0}{2 - 0} = \frac{1}{2}
\]

Thus,

\[
\lim_{x \to \infty} \frac{x + 1}{2x - 3} = \frac{1}{2}
\]

### 2. Limit as \( x \to -\infty \):
Now we compute the limit as \( x \) approaches negative infinity:

\[
\lim_{x \to -\infty} \frac{x + 1}{2x - 3}
\]

Again, dividing both the numerator and the denominator by \( x \):

\[
\lim_{x \to -\infty} \frac{1 + \frac{1}{x}}{2 - \frac{3}{x}} = \frac{1 + 0}{2 - 0} = \frac{1}{2}
\]

Thus,

\[
\lim_{x \to -\infty} \frac{x + 1}{2x - 3} = \frac{1}{2}
\]

### 3. Limit as \( x \to \frac{3}{2} \) (vertical asymptote):
The denominator \( 2x - 3 = 0 \) when \( x = \frac{3}{2} \). To investigate the behavior of the function near this point, we consider the left-hand and right-hand limits as \( x \to \frac{3}{2} \):

#### Left-hand limit:
\[
\lim_{x \to \frac{3}{2}^-} \frac{x + 1}{2x - 3}
\]
As \( x \) approaches \( \frac{3}{2} \) from the left, the denominator becomes small and negative, causing the whole expression to approach \( -\infty \).

#### Right-hand limit:
\[
\lim_{x \to \frac{3}{2}^+} \frac{x + 1}{2x - 3}
\]
As \( x \) approaches \( \frac{3}{2} \) from the right, the denominator becomes small and positive, causing the expression to approach \( +\infty \).

Thus, there is a vertical asymptote at \( x = \frac{3}{2} \), and the function tends to \( \pm\infty \) near this point.

### Summary:
- \( \lim_{x \to \infty} \frac{x + 1}{2x - 3} = \frac{1}{2} \)
- \( \lim_{x \to -\infty} \frac{x + 1}{2x - 3} = \frac{1}{2} \)
- There is a vertical asymptote at \( x = \frac{3}{2} \), and the function approaches \( \pm\infty \) near this point.

Would you like to explore further details or have any questions?

Here are 5 related questions:
1. How do limits help in identifying horizontal asymptotes?
2. What happens to the function when the denominator of a rational function tends to zero?
3. How can you determine if a function has a hole or a vertical asymptote?
4. Can limits be used to study the behavior of functions near infinity?
5. How do left-hand and right-hand limits differ?

**Tip:** When studying limits, always consider both the behavior as \( x \) approaches infinity and as it approaches any critical points where the denominator becomes zero.

By using the theory of limits solve this: y = (x + 1) / (2x - 3)

Learn how to solve the rational function y = (x + 1) / (2x - 3) using the theory of limits. This solution explores limits at infinity, negative infinity, and the vertical asymptote. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation