Problem 54
Question
Find the limits in Exercises \(53-58.\) $$\lim \frac{1}{x^{2}-1}\text{ as }$$ \begin{equation}\begin{array}{ll}{\text { a. }} & {x \rightarrow 1^{+}} & {\text { b. } x \rightarrow 1^{-}} \\ {\text { c. }} & {x \rightarrow-1^{+}} & {\text { d. } x \rightarrow-1^{-}}\end{array}\end{equation}
Step-by-Step Solution
Verified Answer
a. \(+\infty\); b. \(-\infty\); c. \(-\infty\); d. \(+\infty\).
1Step 1: Understand the Function and Limits
First, we consider the function \(f(x) = \frac{1}{x^2 - 1}\). We need to find the values of the function as \(x\) approaches \(1^+\), \(1^-\), \(-1^+\), and \(-1^-\). Note that \(x^2 - 1 = (x-1)(x+1)\), and the function is undefined at \(x = 1\) and \(x = -1\) since the denominator is zero at these points.
2Step 2: Limit as x Approaches 1 from the Right
For \(x \rightarrow 1^+\), \(x\) is slightly greater than 1. In this case, \(x^2 - 1\) is positive because it is the product of a small positive number (\(x-1\)) and the positive number \(2\) (since \(x \approx 1\)). Consequently, \(\frac{1}{x^2 - 1}\) approaches a large positive value. Thus, \(\lim_{{x \to 1^+}} \frac{1}{x^2 - 1} = +\infty\).
3Step 3: Limit as x Approaches 1 from the Left
For \(x \rightarrow 1^-\), \(x\) is slightly less than 1. Here, \(x^2 - 1\) is negative because it's the product of a small negative number (\(x-1\)) and the positive number \(2\). Thus, \(\frac{1}{x^2 - 1}\) approaches a large negative value. So, \(\lim_{{x \to 1^-}} \frac{1}{x^2 - 1} = -\infty\).
4Step 4: Limit as x Approaches -1 from the Right
For \(x \rightarrow -1^+\), \(x\) is slightly greater than -1. In this scenario, \(x^2 - 1\) is negative because \(x+1\) is a small positive number and \(x-1\) is a negative number. Thus, \(\frac{1}{x^2 - 1}\) approaches a large negative value. Therefore, \(\lim_{{x \to -1^+}} \frac{1}{x^2 - 1} = -\infty\).
5Step 5: Limit as x Approaches -1 from the Left
For \(x \rightarrow -1^-\), \(x\) is slightly less than -1. In this case, \(x^2 - 1\) is positive because both \(x+1\) and \(x-1\) are negative, making their product positive. Consequently, \(\frac{1}{x^2 - 1}\) approaches a large positive value. Thus, \(\lim_{{x \to -1^-}} \frac{1}{x^2 - 1} = +\infty\).
Key Concepts
Limit from the RightLimit from the LeftFunction Undefined Points
Limit from the Right
When evaluating a limit from the right, which is written as \( x \to a^+ \), we consider the behavior of the function as \( x \) approaches \( a \) from values greater than \( a \). This means we are looking at values that are just a tiny bit larger than \( a \). The mathematical expression for this is \( \lim_{{x \to a^+}} f(x) \).
Consider the function \( f(x) = \frac{1}{x^2 - 1} \) and the limit as \( x \to 1^+ \). Here, \( x^2 - 1 \) becomes a small positive value since \( x \) is slightly more than 1. The denominator \((x-1)(x+1)\) becomes a positive number because \( x-1 \) is slightly positive and \( x+1 \) is greater than zero. Therefore, as the denominator gets closer to zero from a positive side, the fraction \( \frac{1}{x^2 - 1} \) grows very large positively. This results in the limit being \( +\infty \).
This process illustrates how as we approach a number from the right, the sign and the behavior of the function can be predicted by closely analyzing the contributing factors in the expression.
Consider the function \( f(x) = \frac{1}{x^2 - 1} \) and the limit as \( x \to 1^+ \). Here, \( x^2 - 1 \) becomes a small positive value since \( x \) is slightly more than 1. The denominator \((x-1)(x+1)\) becomes a positive number because \( x-1 \) is slightly positive and \( x+1 \) is greater than zero. Therefore, as the denominator gets closer to zero from a positive side, the fraction \( \frac{1}{x^2 - 1} \) grows very large positively. This results in the limit being \( +\infty \).
This process illustrates how as we approach a number from the right, the sign and the behavior of the function can be predicted by closely analyzing the contributing factors in the expression.
Limit from the Left
The limit from the left, expressed as \( x \to a^- \), examines how a function behaves as \( x \) approaches \( a \) from values that are a tad less than \( a \). It's the limit of the function as \( x \) comes from the left of a specified point.
For instance, consider \( f(x) = \frac{1}{x^2 - 1} \) as \( x \to 1^- \). As \( x \) is slightly less than 1, \( x^2 - 1 \) becomes a small negative number. The factor \((x-1)\) is slightly negative, while \((x+1)\) is positive, making their product a negative number. Because of this negative denominator, the fraction approaches a large negative value when \( x \to 1^- \), leading to the limit \( -\infty \).
Observing from the left helps us determine when the function trends downward dramatically, affected by an inverse proportion in physics and engineering scenarios.
For instance, consider \( f(x) = \frac{1}{x^2 - 1} \) as \( x \to 1^- \). As \( x \) is slightly less than 1, \( x^2 - 1 \) becomes a small negative number. The factor \((x-1)\) is slightly negative, while \((x+1)\) is positive, making their product a negative number. Because of this negative denominator, the fraction approaches a large negative value when \( x \to 1^- \), leading to the limit \( -\infty \).
Observing from the left helps us determine when the function trends downward dramatically, affected by an inverse proportion in physics and engineering scenarios.
Function Undefined Points
Function undefined points occur where the denominator in a function's fraction equals zero, making it impossible to calculate a valid result. Such points need special consideration when finding limits, as they can cause the function to head towards infinity or be classified as an undefined or discontinuous point.
Take, for example, the function \( f(x) = \frac{1}{x^2 - 1} \). The function is undefined at \( x = 1 \) and \( x = -1 \) due to division by zero. Both \( x^2 - 1 \) equals zero at these points, thereby causing vertical asymptotes in the graph of the function.
However, when approaching these undefined points from either direction, we can analyze how the function behaves, such as increasing towards \( +\infty \) or decreasing towards \( -\infty \). Recognizing these undefined points and computing limits around them is crucial when describing the function's behavior near discontinuities.
Take, for example, the function \( f(x) = \frac{1}{x^2 - 1} \). The function is undefined at \( x = 1 \) and \( x = -1 \) due to division by zero. Both \( x^2 - 1 \) equals zero at these points, thereby causing vertical asymptotes in the graph of the function.
However, when approaching these undefined points from either direction, we can analyze how the function behaves, such as increasing towards \( +\infty \) or decreasing towards \( -\infty \). Recognizing these undefined points and computing limits around them is crucial when describing the function's behavior near discontinuities.
Other exercises in this chapter
Problem 53
Suppose \(\lim _{x \rightarrow c} f(x)=5\) and \(\lim _{x \rightarrow g} g(x)=-2 .\) Find \begin{equation} \text { a. }\lim _{x \rightarrow c} f(x) g(x) \quad \
View solution Problem 54
A function value Show that the function \(F(x)=(x-a)^{2}\) . \((x-b)^{2}+x\) takes on the value \((a+b) / 2\) for some value of \(x .\)
View solution Problem 54
Suppose \(\lim _{x \rightarrow 4} f(x)=0\) and \(\lim _{x \rightarrow 4} g(x)=-3 .\) Find \begin{equation} \text {a. }\lim _{x \rightarrow 4}(g(x)+3) \quad \tex
View solution Problem 54
Another wrong statement about limits Show by example that the following statement is wrong. $$ \begin{array}{l}{\text { The number } L \text { is the limit of }
View solution