Problem 59 - Calculus and its applications

Question

Graph each function and then find the specified limits. When necessary, state that the limit does not exist. $$ g(x)=\frac{1}{x-3}+2 ; \quad \text { find } \lim _{x \rightarrow \infty} g(x) \text { and } \lim _{x \rightarrow 3} g(x). $$

Step-by-Step Solution

Answer

\( \lim_{x \to \infty} g(x) = 2 \), \( \lim_{x \to 3} g(x) \) does not exist.

Step 1: Understand the Function

The function given is a transformation of the basic rational function \( g(x) = \frac{1}{x-3} + 2 \). This translates the basic function \( f(x) = \frac{1}{x} \) to the right by 3 units and up by 2 units.

Step 2: Graph the Function

To graph \( g(x) = \frac{1}{x-3} + 2 \), first note the vertical asymptote at \( x = 3 \) and the horizontal asymptote at \( y = 2 \). The graph will approach these asymptotes but never touch them. Sketch the curve accordingly, taking into account these shifts and asymptotes.

Step 3: Evaluate \( \lim_{x \to \infty} g(x) \)

For \( x \to \infty \), the term \( \frac{1}{x-3} \) approaches 0 because the denominator becomes very large. Thus, \( g(x) \to 0 + 2 \). Therefore, \( \lim_{x \to \infty} g(x) = 2 \).

Step 4: Evaluate \( \lim_{x \to 3} g(x) \)

As \( x \) approaches 3, the denominator \( x-3 \) of the first term approaches 0, which makes \( \frac{1}{x-3} \) tend toward \( \pm \infty \), depending on the direction of approach. Consequently, \( \lim_{x \to 3} g(x) \) does not exist because of the vertical asymptote at \( x = 3 \).

Key Concepts

Rational Functions

Rational functions are fractions composed of polynomial expressions in the numerator and the denominator. In a more straightforward sense, they look like ratios of two polynomial functions. The most basic example is \[ f(x) = \frac{1}{x} \].
These functions are quite intriguing because they can exhibit unique behaviors that simple polynomials cannot. Unlike polynomial functions, rational functions can have restrictions on their domains. Therefore, you can only input values that do not make the denominator zero.
An important aspect of rational functions is that they may have asymptotes, lines that the graph approaches but never actually touches, as seen in the example function \( g(x) = \frac{1}{x-3} + 2 \). These asymptotes significantly impact the graph's overall shape and behavior.

Asymptotes

Asymptotes are lines that a graph approaches but never touches or crosses. They help describe the long-term behavior of a function. Asymptotes can be vertical, horizontal, or even oblique.
In our function \( g(x) = \frac{1}{x-3} + 2 \), we've got:

• Vertical Asymptote: When you set the denominator equal to zero, the graph tends towards infinity, creating a vertical asymptote. For \( g(x) \), this occurs at \( x = 3 \).
• Horizontal Asymptote: As \( x \) approaches infinity or negative infinity in rational functions with no overall dominant term, the function approaches a constant value, forming a horizontal asymptote. Here, as \( x \rightarrow \infty, \ g(x) \rightarrow 2 \).

These act like invisible boundaries that show how the graph behaves as it heads into extreme values.

Graphing Functions

Graphing rational functions involves determining their key features: asymptotes, intercepts, and the general shape of the curve. To properly graph \( g(x) = \frac{1}{x-3} + 2 \), one should:

• Identify vertical asymptotes by setting the denominator equal to zero and solving for \( x \).
• Determine horizontal asymptotes by examining the limits as \( x \) approaches infinity. For most cases, you look at the leading coefficient of the numerator divided by the leading coefficient of the denominator.
• Plot additional points to accurately reflect the curve. Select a range of \( x \) values, especially close to the asymptotes, to understand the graph's behavior.

Graphing functions like this helps visualize the behavior and relationships, explaining more about limits and the function's infinity behaviors, which are often unseen with mere equations.