Problem 30 - Calculus Early Transcendentals

Question

Analyze the following limits and find the vertical asymptotes of \(f(x)=\frac{x+7}{x^{4}-49 x^{2}}\) a. \(\lim _{x \rightarrow 7} f(x)\) b. \(\lim _{x \rightarrow 7^{+}} f(x)\) \(\lim _{x \rightarrow-7} f(x) \quad\) d. \(\lim _{x \rightarrow 0} f(x)\)

Step-by-Step Solution

Answer

a) The limit as \(x\) approaches 7 does not exist. b) The limit as \(x\) approaches 7 from the right is +∞. c) The limit as \(x\) approaches -7 does not exist. d) The limit as \(x\) approaches 0 is -∞.

Step 1: Identifying Vertical Asymptotes

A vertical asymptote occurs where the function is undefined, which happens when the denominator of the function equals zero. So, let's set the denominator equal to zero and solve for x: \(x^4 - 49x^2 = 0\) \(x^2(x^2 - 49) = 0\) \(x^2 = 0\quad\) or \(\quad x^2 - 49 = 0\) \(x = 0\quad\) or \(\quad x^2 = 49\) \(x = 0\quad\) or \(\quad x = \pm7\) So, we have vertical asymptotes at \(x = -7, x = 0,\) and \(x = 7\). Now, we can find the specified limits while considering these values of x.

Step 2: Finding the Limit as \(x\) approaches 7

a. \(\lim _{x \rightarrow 7} f(x)\) Since there is a vertical asymptote at \(x = 7\), the function is undefined at this point, and we cannot directly substitute the value into the function. Thus, this limit does not exist.

Step 3: Finding the Limit as \(x\) approaches 7 from the right

b. \(\lim _{x \rightarrow 7^{+}} f(x)\) As x approaches 7 from the right, we consider the behavior of the function near the vertical asymptote at \(x = 7\). Notice that both the numerator and denominator are positive for \(x>7\). Therefore, this limit has a positive infinity as the value, meaning the graph of the function increases without bound: \(\lim _{x \rightarrow 7^{+}} f(x) = +\infty\)

Step 4: Finding the Limit as \(x\) approaches \(-7\)

c. \(\lim _{x \rightarrow -7} f(x)\) Similar to the case of \(\lim _{x \rightarrow 7} f(x)\), there is a vertical asymptote at \(x = -7\). Therefore, the function is undefined at this point, and this limit does not exist.

Step 5: Finding the Limit as \(x\) approaches 0

d. \(\lim _{x \rightarrow 0} f(x)\) At \(x = 0\), the vertical asymptote exists. Given this information, we can determine the behavior of the function as \(x\) approaches 0. It's important to note that the numerator will be positive (since there is a constant \(+7\)) and the denominator will be negative (due to the x values being squared). Therefore, \(\lim _{x \rightarrow 0} f(x) = -\infty\). To summarize: a. \(\lim _{x \rightarrow 7} f(x)\) does not exist b. \(\lim _{x \rightarrow 7^{+}} f(x) = +\infty\) c. \(\lim _{x \rightarrow -7} f(x)\) does not exist d. \(\lim _{x \rightarrow 0} f(x) = -\infty\)

Key Concepts

Limits Analysis

Understanding limits is crucial in calculus, especially when analyzing how functions behave as they approach specific points. A limit explores the function's value as the input approaches a given point, not necessarily at that point.

When carrying out a limits analysis, we investigate the function's behavior around points of interest, which could be real numbers or infinity. This analysis is pivotal in identifying discontinuities, including vertical asymptotes. It involves looking from both sides of the point (from the left and the right) to see if the function approaches the same value or diverges.

For instance, in analyzing the limits of the function \(f(x)=\frac{x+7}{x^4-49x^2}\), we check what happens as x approaches 7, -7 and 0. Our analysis concludes that the function does not settle at any finite value at these points. Instead, it either does not exist or reaches positive or negative infinity, indicative of a vertical asymptote.

Undefined Function Points

A function is undefined at points where it does not produce a valid output. In the context of rational functions like \(f(x)=\frac{x+7}{x^4-49x^2}\), undefined points occur when the denominator equals zero, since division by zero is not permissible in mathematics.

Identifying these undefined function points can be done by setting the denominator equal to zero and solving for possible x-values. As shown in the solution, \(x^2(x^2 - 49) = 0\) reveals that x equals 0, 7, and -7 are points where the function \(f(x)\) does not produce a valid output and thus is undefined. These points often correspond to vertical asymptotes in the function's graph, marking locations where the function shoots off towards infinity.

Limit Behavior Near Asymptotes

Vertical asymptotes reflect dramatic changes in a function's behavior, showing where a function tends towards positive or negative infinity. But just knowing where the asymptotes are isn't enough; how the function approaches the asymptote—that is, its limit behavior—is equally significant.

As x approaches the vertical asymptote, the function's value typically grows without bound in the positive or negative direction. The specific behavior can be assessed using one-sided limits. For example, as \(x\) approaches 7 from the right in \(f(x)\), the function increases without limit, hence the \(\lim _{x \rightarrow 7^{+}} f(x) = +\infty\). On the other hand, as \(x\) approaches 0, the function dives without limit, so \(\lim _{x \rightarrow 0} f(x) = -\infty\). These limits do not yield a particular value but indicate that the function is not approaching a fixed value as it nears the asymptote.