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Chapter 2: Problem 28

Decide which of the given one-sided or two-sided limits exist as numbers, which as \(\infty\), which as \(-\infty\), and which do not exist. Where the limit is a number, evaluate it. $$ \lim _{x \rightarrow-1^{+}} \frac{x^{2}+5 x+4}{x+1} $$

Short Answer

Expert verified
The limit exists as a number and equals 3.

Step by step solution

01

Understand the problem

We are given the limit \( \lim _{x \rightarrow-1^{+}} \frac{x^{2}+5x+4}{x+1} \). We need to evaluate this one-sided limit as \( x \) approaches \(-1\) from the right side.
02

Identify potential issues

As \( x \to -1^{+} \), the denominator \( x + 1 \to 0 \). This signals a potential vertical asymptote or hole at \( x = -1 \). Our goal is to see if the numerator also approaches 0, which would indicate a removable discontinuity.
03

Factor the numerator

Factor the numerator \( x^2 + 5x + 4 \) into \((x + 1)(x + 4)\). The expression becomes \( \frac{(x + 1)(x + 4)}{x + 1} \).
04

Simplify the expression

Cancel the common factor \((x + 1)\) from the numerator and the denominator, giving us \( x + 4 \). This indicates that the discontinuity at \( x = -1 \) is removable.
05

Evaluate the limit

Now, with the simplified expression \( x + 4 \), substitute \( x = -1 \): \[(x + 4)_{x = -1} = -1 + 4 = 3.\]
06

Determine the existence and value of the limit

Since the expression was simplified to be continuous at \( x = -1 \) after cancellation, the limit exists as a number. The value of the limit as \( x \to -1^{+} \) is 3.

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Key Concepts

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

One-Sided Limits
One-sided limits focus on understanding how a function behaves as a variable approaches a certain point from one side only. In our exercise, we analyzed the limit \( \lim_{x \rightarrow -1^{+}} \frac{x^{2}+5x+4}{x+1} \). The symbol \( -1^{+} \) indicates that we are observing the behavior of the function as \( x \) approaches \(-1\) from the right-hand side. This concept is crucial because the behavior from one side can differ drastically compared to the other.
  • One-sided limits are different from two-sided limits because they only consider one direction of approach.
  • They help in identifying the existence of vertical asymptotes, where the function diverges to \( \infty \) or \( -\infty \) from one side.
  • It is especially useful for functions with discontinuities.
By only considering the right-hand approach, we notice how variables behave similarly or differently compared to approaching from the left.
Factorization
Factorization helps in simplifying expressions, making it easier to evaluate limits. In the provided solution, the numerator \( x^2 + 5x + 4 \) was factored into \((x+1)(x+4)\). This step was crucial to understand the limit's behavior further.
  • Factorization involves breaking down expressions into products of simpler factors.
  • This can help identify common terms in the numerator and denominator that might cancel out, simplifying the expression.
  • In this exercise, factorization revealed a removable discontinuity.
This process often transforms complex rational expressions into simpler forms, providing clarity on the behavior near points of interest.
Removable Discontinuity
A removable discontinuity is a point where a function is not defined due to a factor that can be canceled out, such as \( x + 1 \) in both the numerator and denominator in our example. After cancellation, the expression becomes \( x+4 \), whose behavior is simple to analyze.
  • It occurs where the function seems to "break" but can be "patched" by defining the point with a new value.
  • After removing the discontinuity, the function becomes continuous at that point.
  • Understanding removable discontinuities is key in evaluating limits where division by zero might initially seem an issue.
By recognizing and removing this discontinuity, we established the limit's existence and evaluated it easily, which in our solution, led us to the limit 3.
Vertical Asymptotes
Vertical asymptotes are lines \( x = a \) where a function tends to \( \pm \infty \) as \( x \) approaches \( a \), often occurring when the denominator of a rational expression approaches zero. In this exercise, analyzing \( x + 1 \to 0 \) upon approaching \( x = -1 \) initially suggested a potential asymptote.
  • Vertical asymptotes occur due to division by zero circumstances in rational functions, often indicating non-removable discontinuities.
  • On either side of the asymptote, the function will diverge drastically, making one-sided analysis useful.
  • Not all points where the denominator is zero become asymptotes; factorization can reveal removable discontinuities instead.
In our solution, factorization showed a removable discontinuity rather than an asymptote, confirming the limit as \( x \rightarrow -1^{+} \) exists and is 3.

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Most popular questions from this chapter

Suppose \(\lim _{x \rightarrow a} f(x)=-2\) and \(\lim _{x \rightarrow a} g(x)=3\). a. Guess the value of \(\lim _{x \rightarrow a}[f(x) g(x)]\). b. Guess the value of \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\).

Explain why \(f\) is continuous on the given interval or intervals $$ f(x)=\sqrt{x+3} ;[-3, \infty) $$

Evaluate the function at \(0.1,0.01\), and \(0.001\), and at \(-0.1,-0.01\), and \(-0.001\). Then guess the value of \(\lim _{x \rightarrow 0} f(x) .\) \(f(x)=\frac{\ln (1+x)}{x}\)

Show that the equation has at least one solution. $$ x^{2}+\frac{1}{x}=1 $$

Decide which of the given one-sided or twosided limits exist as numbers, which as \(\infty\), which as \(-\infty\), and which do not exist. Where the limit is a number, evaluate it. $$ \lim _{x \rightarrow 0} \sqrt{\frac{1}{x^{2}}} $$
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