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Chapter 3: Problem 12

Evaluate the limits in problems. $$ \lim _{x \rightarrow-\infty} \frac{2 x+x^{2}}{3 x+1} $$

Short Answer

Expert verified
The limit is \(-\infty\).

Step by step solution

01

Identify the Highest Power of x

In the given expression \( \frac{2x + x^2}{3x + 1} \), identify the highest power of \( x \) in both the numerator and the denominator. Here, the highest power in the numerator is \( x^2 \), and in the denominator, it is \( x \).
02

Divide Every Term by the Highest Power in the Denominator

Divide each term in the numerator and denominator by \( x \), the highest power found in the denominator. The expression becomes \( \lim_{x \to -\infty} \frac{\frac{2x}{x} + \frac{x^2}{x}}{\frac{3x}{x} + \frac{1}{x}} \).
03

Simplify the Expression

Simplifying, we have: \( \lim_{x \to -\infty} \frac{2 + x}{3 + \frac{1}{x}} \). Now, consider the behavior of each term as \( x \to -\infty \). The term \( \frac{1}{x} \) approaches 0.
04

Evaluate the Limit

Substituting the values as \( x \to -\infty \), in the simplified limit, the expression becomes \( \frac{2 + (-\infty)}{3} = \frac{-\infty}{3} \). This evaluates to \( -\infty \).

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Key Concepts

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

Understanding the Highest Power of x
In problems involving limits, especially with rational functions, identifying the highest power of \( x \) is crucial. Let鈥檚 look at the example of \( \frac{2x + x^2}{3x + 1} \). In this expression, we must recognize which terms dominate the behavior of the function as \( x \) becomes very large or very small. Here鈥檚 the key idea:
  • The highest power of \( x \) in the numerator is \( x^2 \).
  • The highest power of \( x \) in the denominator is \( x \).
Why does this matter? Because as \( x \) approaches infinity or negative infinity, terms with lower powers of \( x \) grow insignificant compared to those with the highest power. This helps simplify the process of finding limits by focusing on dominant terms in both the numerator and the denominator.
Rational Functions and Their Limits
Rational functions consist of a polynomial in the numerator and another in the denominator. Understanding their limit behavior is about simplifying these functions intelligently.
To identify a rational function's limit as \( x \) approaches a very large number (positive or negative infinity), we follow these steps:
  • Determine the highest power of \( x \) in both the numerator and denominator.
  • Divide every term by the highest power found in the denominator.
By simplifying the expression \( \frac{2x + x^2}{3x + 1} \) as \( x \to -\infty \), the expression becomes \( \lim_{x \to -\infty} \frac{2 + x}{3 + \frac{1}{x}} \). Here, you can see that the expression's limit largely depends on the highest power terms. This is a streamlined approach, making complex calculations more manageable.
Behavior as x Approaches Infinity
Exploring the behavior as \( x \) approaches infinity or negative infinity is all about knowing which terms affect the limit鈥檚 value. In our expression \( \lim_{x \to -\infty} \frac{2 + x}{3 + \frac{1}{x}} \), each term's behavior matters:
  • The term \( x \) in the numerator becomes dominant as it trends towards \(-\infty\), overpowering the constant \( 2 \).
  • In the denominator, the term \( \frac{1}{x} \) approaches \(0\), making it negligent.
Given these behaviors, when you substitute the infinite values, \( \frac{2 + (-\infty)}{3} \) results in \( \frac{-\infty}{3} \), simplifying to \(-\infty\). This illustrates how larger powers drive the behavior, influencing the resulting limit significantly.

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Most popular questions from this chapter

In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow-2} \sqrt{6+x} $$

In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 2^{+}} \frac{2}{x^{2}-4} $$

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin x}{-x} $$

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{1-\cos (5 x)}{2 x} $$

(a) Use a graphing calculator to sketch the graph of $$f(x)=e^{a x} \sin x, \quad x \geq 0$$ for \(a=-0.1,-0.01,0,0.01\), and \(0.1\). (b) Which part of the function \(f(x)\) produces the oscillations that you see in the graphs sketched in (a)? (c) Describe in words the effect that the value of \(a\) has on the shape of the graph of \(f(x)\) (d) Graph \(f(x)=e^{a x} \sin x, g(x)=-e^{a x}\), and \(h(x)=e^{a x}\) together in one coordinate system for (i) \(a=0.1\) and (ii) \(a=\) \(-0.1 .\) [Use separate coordinate systems for (i) and (ii).] Explain what you see in each case. Show that $$-e^{a x} \leq e^{a x} \sin x \leq e^{a x}$$ Use this pair of inequalities to determine the values of \(a\) for which $$\lim _{x \rightarrow \infty} f(x)$$ exists, and find the limiting value.
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