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Chapter 2: Problem 24

Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{x \rightarrow-3} \frac{x+3}{x^{2}+4 x+3}$$

Short Answer

Expert verified
The limit is \(-\frac{1}{2}\).

Step by step solution

01

Direct Substitution

To find the limit \( \lim_{x \to -3} \frac{x+3}{x^2+4x+3} \), the first step is to perform a direct substitution of \( x = -3 \) into the function. Substitute \( x = -3 \) into \( x + 3 \) and \( x^2 + 4x + 3 \):\[ f(-3) = \frac{-3 + 3}{(-3)^2 + 4(-3) + 3} = \frac{0}{9 - 12 + 3} = \frac{0}{0}. \]This results in an indeterminate form \( \frac{0}{0} \), so direct substitution cannot provide the limit; further analysis is needed.
02

Factor the Expression

Given that direct substitution yields \( \frac{0}{0} \), we need to simplify the expression by factoring. Factor both the numerator and the denominator:1. Numerator: \( x + 3 \) is already a simple factor.2. Denominator: The quadratic expression can be factored:\[ x^2 + 4x + 3 = (x + 3)(x + 1). \]This gives the expression:\[ \frac{x + 3}{(x + 3)(x + 1)}. \]
03

Simplify the Expression

Now, simplify the expression by canceling common factors from the numerator and the denominator:\[ \frac{x + 3}{(x + 3)(x + 1)} = \frac{1}{x + 1}, \]provided \( x eq -3 \) to avoid division by zero. With this simplification, the function is easier to analyze.
04

Evaluate the Limit

Since the expression has been simplified to \( \frac{1}{x + 1} \), we can now substitute \( x = -3 \) directly into this simplified expression:\[ \lim_{x \to -3} \frac{1}{x + 1} = \frac{1}{-3 + 1} = \frac{1}{-2} = -\frac{1}{2}. \]

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Key Concepts

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

Direct Substitution
Direct substitution is typically the first step when trying to determine the limit of a function. The aim is to substitute the given value directly into the function. In this exercise, we attempted to substitute \(x = -3\) into \(\lim_{x \to -3} \frac{x+3}{x^2+4x+3}\). When substituted directly, this yields
  • Numerator: \(-3 + 3 = 0\)
  • Denominator: \( (-3)^2 + 4(-3) + 3= 9 - 12 + 3 = 0\)
This evaluation resulted in the form \(\frac{0}{0}\), known as an indeterminate form. Direct substitution alone doesn't solve the limit here because of this form. This step indicates a need for alternative methods to find the limit.
Factorization
Factorization is a technique used to simplify expressions, especially when dealing with indeterminate forms. In our exercise, the challenge was to simplify \(\frac{0}{0}\) to find a limit.
  • The numerator \(x + 3\) is already in its simplest form.
  • The quadratic expression in the denominator, \(x^2 + 4x + 3\), can be factored into \((x + 3)(x + 1)\).
Factoring the denominator reveals common factors with the numerator. Now, we can cancel the common term \(x + 3\). After cancellation, the expression becomes \(\frac{1}{x + 1}\), provided \(x eq -3\), as division by zero is undefined. This step significantly simplifies the function and is key to evaluating the limit that follows.
Indeterminate Forms
Indeterminate forms arise in calculus when direct substitution into a limit yields results like \(\frac{0}{0}\). These forms are significant because they indicate that further analysis is required to evaluate the limit. When encountering an indeterminate form, you can:
  • Factorize: Simplify the function by finding and canceling common factors.
  • Use L'H么pital's Rule: An advanced technique when differentiation applies, though not used in this exercise.
  • Other algebraic manipulations: Such as multiplying by a conjugate or rationalizing.
In this exercise, our initial attempt through direct substitution resulted in the indeterminate form \(\frac{0}{0}\). By recognizing this form, the factorization step becomes necessary. This approach simplifies the function making it easier to reevaluate the limit successfully.

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Most popular questions from this chapter

In Exercises \(73-76,\) find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$\lim _{x \rightarrow-\infty} h(x)=-1, \lim _{x \rightarrow \infty} h(x)=1, \lim _{x \rightarrow 0^{-}} h(x)=-1,\text{and}\\\\{\lim _{x \rightarrow 0^{+}} h(x)=1}$$

You will find a graphing calculator useful. Let $$h(x)=\left(x^{2}-2 x-3\right) /\left(x^{2}-4 x+3\right)$$ a. Make a table of the values of \(h\) at \(x=2.9,2.99,2.999,\) and so on. Then estimate lim_{x} \rightarrow 3 \(h(x)\) . What estimate do you arrive at if you evaluate \(h\) at \(x=3.1,3.01,3.001, \ldots\) instead? b. Support your conclusions in part (a) by graphing \(h\) near \(x_{0}=3\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow 3\) . c. Find lim_{x\rightarrow3} \(h(x)\) algebraically.

Use a CAS to perform the following steps: a. Plot the function near the point \(x_{0}\) being approached. b. From your plot guess the value of the limit. $$\lim _{x \rightarrow 0} \frac{2 x^{2}}{3-3 \cos x}$$

Roots of a cubic Show that the equation \(x^{3}-15 x+1=0\) has three solutions in the interval \([-4,4] .\)

Removable discontinuity Give an example of a function \(f(x)\) that is continuous for all values of \(x\) except \(x=2,\) where it has a removable discontinuity. Explain how you know that \(f\) is discontinuous at \(x=2,\) and how you know the discontinuity is removable. A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function $$ f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right. $$ is discontinuous at every point. b. Is \(f\) right-continuous or left-continuous at any point?
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