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Chapter 2: Problem 53

Find the limits. \(\lim \frac{1}{x^{2}-4}\) as a. \(x \rightarrow 2^{+}\) b. \(x \rightarrow 2^{-}\) c. \(x \rightarrow-2^{+}\) d. \(x \rightarrow-2^{-}\)

Short Answer

Expert verified
a. \( +\infty \); b. \( -\infty \); c. \( -\infty \); d. \( +\infty \).

Step by step solution

01

Identify the Problem

We need to find the limit of \( \frac{1}{x^2-4} \) as \( x \) approaches 2 from the right (\( x \to 2^+ \)), 2 from the left (\( x \to 2^- \)), -2 from the right (\( x \to -2^+ \)), and -2 from the left (\( x \to -2^- \)).
02

Rewriting the Function

The function \( \frac{1}{x^2 - 4} \) can be rewritten as \( \frac{1}{(x-2)(x+2)} \). This will help in evaluating the behavior as \( x \) approaches 卤2.
03

Analyzing Limit as \( x \to 2^+ \)

As \( x \to 2^+ \), \( x - 2 \to 0^+ \) and \( x + 2 \to 4 \). Therefore, \( \frac{1}{(x-2)(x+2)} \) becomes \( \frac{1}{0^+ \cdot 4} \), which tends to \( +\infty \).
04

Analyzing Limit as \( x \to 2^- \)

As \( x \to 2^- \), \( x - 2 \to 0^- \) and \( x + 2 \to 4 \). Therefore, \( \frac{1}{(x-2)(x+2)} \) becomes \( \frac{1}{0^- \cdot 4} \), which tends to \( -\infty \).
05

Analyzing Limit as \( x \to -2^+ \)

As \( x \to -2^+ \), \( x + 2 \to 0^+ \) and \( x - 2 \to -4 \). Therefore, \( \frac{1}{(x-2)(x+2)} \) becomes \( \frac{1}{-4 \cdot 0^+} \), which tends to \( -\infty \).
06

Analyzing Limit as \( x \to -2^- \)

As \( x \to -2^- \), \( x + 2 \to 0^- \) and \( x - 2 \to -4 \). Therefore, \( \frac{1}{(x-2)(x+2)} \) becomes \( \frac{1}{-4 \cdot 0^-} \), which tends to \( +\infty \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
When we talk about continuity in mathematics, we are referring to how smoothly a function behaves over its domain. A function is continuous at a point if its graph doesn't have any breaks, holes, or gaps at that point.
To check for continuity at a point, we need to ensure the following:
  • The function is defined at the point.
  • The limit exists as it approaches the point from both sides.
  • The function's value equals the limit.
In the exercise with the function \( f(x) = \frac{1}{x^2-4} \), observe what happens as \( x \) approaches 2 and -2. At these points, the function is not continuous because it becomes undefined (division by zero potential).
Since the limits are infinite and don't match the function's value (as it doesn't exist), the function is discontinuous at these points.
One-sided Limits
One-sided limits help us understand the behavior of a function as it approaches a particular point from one side only, either the left or the right.
For a limit \( \lim_{x \to c^+} f(x) \), we consider the values of \( f(x) \) as \( x \) gets closer to \( c \) from the right.
For a limit \( \lim_{x \to c^-} f(x) \), we consider the values as \( x \) gets closer from the left.
  • In the exercise for \( x \to 2^+ \), \( \lim \frac{1}{x^2 - 4} = +\infty \).
  • For \( x \to 2^- \), the limit is \( -\infty \).
  • For \( x \to -2^+ \), it approaches \( -\infty \).
  • And for \( x \to -2^- \), the limit is \( +\infty \).
One-sided limits reveal that on each side of 2 and -2, the function behaves very differently, making these points of discontinuity.
Infinite Limits
In mathematics, infinite limits occur when the values of a function increase or decrease without bound as \( x \) approaches a specific point. In the context of this exercise, as \( x \to 2 \) or \( x \to -2 \), the function \( \frac{1}{x^2 - 4} \) results in an infinite limit because the denominator of the fraction approaches zero.
Here鈥檚 why infinite limits are crucial:
  • They indicate vertical asymptotes at points where the function heads to infinity.
  • They help identify behavior trends that reveal discontinuities.
In your exercise, observe how the limits differ as \( x \) approaches these values from different directions, leading to \( +\infty \) or \( -\infty \). Recognizing infinite limits gives you insight into understanding where extreme values occur in complex functions.

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Most popular questions from this chapter

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=\sqrt{x}, \quad x=7$$

Manufacturing electrical resistors Ohm's law for electrical circuits like the one shown in the accompanying figure states that \(V=R I .\) In this equation, \(V\) is a constant voltage, \(I\) is the current in amperes, and \(R\) is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which \(V\) will be 120 volts and \(I\) is to be \(5 \pm 0.1\) amp. In what interval does \(R\) have to lie for \(I\) to be within 0.1 amp of the value \(I_{0}=5 ?\) GRAPH CANT COPY Showing \(L\) is not a limit We can prove that \(\lim _{x \rightarrow c} f(x) \neq L\) by providing an \(\epsilon>0\) such that no possible \(\delta>0\) satisfies the condition $$\text { for all } x, \quad 0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon $$We accomplish this for our candidate \(\epsilon\) by showing that for each \(\delta>0\) there exists a value of \(x\) such that$$0<|x-c|<\delta \quad \text { and } \quad|f(x)-L| \geq \epsilon$$ GRAPH CANT COPY

To prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. $$x^{x}=2$$

Gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon$$ $$f(x)=3-2 x, \quad c=3, \quad \epsilon=0.02$$

Explain why the following five statements ask for the same information. a. Find the roots of \(f(x)=x^{3}-3 x-1\) b. Find the \(x\) -coordinates of the points where the curve \(y=x^{3}\) crosses the line \(y=3 x+1\) c. Find all the values of \(x\) for which \(x^{3}-3 x=1\) a. Find the \(x\) -coordinates of the points where the cubic curve \(y=x^{3}-3 x\) crosses the line \(y=1\) e. Solve the equation \(x^{3}-3 x-1=0\)
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