Problem 45 - Calculus: Early Transcendentals

Question

Location of vertical asymptotes Analyze the following limits and find the vertical asymptotes of \(f(x)=\frac{x-5}{x^{2}-25}\) a. \(\lim _{x \rightarrow 5} f(x)\) b. \(\lim _{x \rightarrow-5^{-}} f(x)\) c. \(\lim _{x \rightarrow-5^{+}} f(x)\)

Step-by-Step Solution

Answer

Answer: The vertical asymptotes for the function \(f(x) = \frac{x-5}{x^{2}-25}\) are at \(x = 5\) and \(x = -5\).

Step 1: Find the values of x for which the denominator is zero

We first need to find the values of x that will make the denominator equal to zero. To do this, we set the denominator equal to zero and solve for x: \[x^{2}-25 = 0\] We factor the expression to get: \[(x-5)(x+5) = 0\] Thus, the denominator is equal to zero when \(x=5\) or \(x=-5\).

Step 2: Vertical Asymptote at x = 5

Now, let's analyze the vertical asymptote at the first possible point, \(x = 5\). To do this, we will need to find the limit as x approaches 5. We are given the first limit: \[a. \lim _{x \rightarrow 5} f(x)\] Since the numerator is not equal to zero while the denominator is equal to zero at x = 5, the limit will go to infinity or negative infinity: \[\lim _{x \rightarrow 5} \frac{x-5}{x^{2}-25} = \infty\] Thus, there is a vertical asymptote at \(x=5\).

Step 3: Vertical Asymptote at x = -5

Now, let's analyze the vertical asymptote at the second possible point, \(x = -5\). We will need to find the limit as x approaches -5 from both the left and the right, as given limits b and c: \[b. \lim _{x \rightarrow-5^{-}} f(x)\] \[c. \lim _{x \rightarrow-5^{+}} f(x)\] We know that if either of these limits goes to infinity or negative infinity, then there is a vertical asymptote at that point. Let's consider the limit as x approaches -5 from the left: \[\lim _{x \rightarrow -5^{-}} \frac{x-5}{x^{2}-25} = -\infty\] Now, let's consider the limit as x approaches -5 from the right: \[\lim _{x \rightarrow -5^{+}} \frac{x-5}{x^{2}-25} = \infty\] Both limits go to infinity or negative infinity, confirming that there is a vertical asymptote at \(x = -5\).

Step 4: Final Answer

In conclusion, after analyzing the limits, we find that there are vertical asymptotes for the function \(f(x) = \frac{x-5}{x^{2}-25}\) at \(x = 5\) and \(x = -5\).

Key Concepts

Limits

Understanding limits is crucial when analyzing the behavior of mathematical functions, especially when you want to pinpoint vertical asymptotes. A limit evaluates what value a function approaches as the input gets closer to a specified point. If a function approaches an infinite value at any point, this often hints at a vertical asymptote. For example, when evaluating \[ \lim_{x \to 5} \frac{x-5}{x^2-25} \]we realize the denominator equals zero, leading the function's value towards infinity or negative infinity. Consequently, the point creates an infinity wall in the function graph, indicating the presence of a vertical asymptote. Identifying limits at various points is essential for graphing and understanding complex functions.

Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. These functions can exhibit interesting behaviors, such as having vertical asymptotes or holes, depending on the polynomial expressions involved.For the example function \[ f(x) = \frac{x-5}{x^2-25} \],we see a classic form of rational function with a first-degree polynomial in the numerator and a degree-two polynomial in the denominator. Important characteristics, like vertical asymptotes, occur when the denominator of the rational expression equals zero but the numerator does not. Inspecting these zero points can help identify asymptotes and map the function's behavior across different degrees of input values.

Factoring Polynomials

Factoring polynomials plays a crucial role in simplifying expressions and solving equations, especially in the context of rational functions. Identifying solutions begins by setting the polynomial equal to zero and solving for the variable.For example, we can factor the denominator of the rational function \[ x^2-25 \] into \[ (x-5)(x+5) \].This process reveals the critical x-values, specifically where vertical asymptotes might occur, as these are the solutions that cause the denominator to become zero. Factoring helps locate these problematic points, aiding your ability to fully understand and analyze the function's behavior across different regions.