/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Find the limit. $$ \lim _{x ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit. $$ \lim _{x \rightarrow-\infty} \frac{2 x^{3}+x^{2}+3}{x+1} $$

Short Answer

Expert verified
The short answer is: \(\lim _{x \rightarrow-\infty} \frac{2 x^{3}+x^{2}+3}{x+1} = -\infty \).

Step by step solution

01

Identify the highest power of x in the numerator

Observe the given rational function: \(\frac{2x^3+x^2+3}{x+1}\). The highest power of x in the numerator is x^3.
02

Divide each term in the numerator by the highest power of x identified in Step 1 and simplify

Divide each term in the numerator by x^3, rewrite and simplify the given limit as: \[ \lim_{x \rightarrow -\infty} \frac{2 + \frac{1}{x} + \frac{3}{x^3}}{\frac{1}{x^2} + \frac{1}{x^3}} \]
03

Evaluate the limit

Now we will split up the fractions and evaluate the limit for each fraction: \[ \lim_{x \rightarrow -\infty} \frac{2}{\frac{1}{x^2}} + \lim_{x \rightarrow -\infty} \frac{\frac{1}{x}}{\frac{1}{x^2}} + \lim_{x \rightarrow -\infty} \frac{\frac{3}{x^3}}{\frac{1}{x^2}} = 2 \lim_{x \rightarrow -\infty} \frac{1}{\frac{1}{x^2}} + \lim_{x \rightarrow -\infty} x + 3\lim_{x \rightarrow -\infty} \frac{1}{x} \] As x approaches negative infinity, \(\lim_{x \rightarrow -\infty} \frac{1}{x} = 0\), \(\lim_{x \rightarrow -\infty} \frac{1}{x^2} = 0\), and \(\lim_{x \rightarrow -\infty} \frac{1}{x^3} = 0\).
04

Final result

Hence, the value of the limit is: \[ 2 \lim_{x \rightarrow -\infty} \frac{1}{\frac{1}{x^2}} + \lim_{x \rightarrow -\infty} x + 3\lim_{x \rightarrow -\infty} \frac{1}{x} = 2(0) + (-\infty) + 3(0) = -\infty \] Therefore, \(\lim _{x \rightarrow-\infty} \frac{2 x^{3}+x^{2}+3}{x+1} = -\infty \).

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Key Concepts

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

Rational Functions
Rational functions are expressions that are formed by the ratio of two polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials with \( Q(x) \) not equal to zero.
These functions are particularly interesting because they can have values approaching infinity or become undefined at certain points.
In the given exercise, the function \( \frac{2x^3 + x^2 + 3}{x+1} \) is a rational function. Understanding the behavior of these functions is crucial for finding limits, especially limits at infinity, as they capture how the function behaves as \( x \) moves towards infinitely large positive or negative values.
Limits at Infinity
The concept of limits at infinity is about understanding how a function behaves as the variable \( x \) approaches infinity, either positive or negative.
This is important for analyzing the end behavior of a function. For rational functions, the behavior of the function as \( x \to \infty \) or \( x \to -\infty \) often depends on the degrees of the polynomials in the numerator and the denominator.
For the exercise, as \( x \to -\infty \), we looked at the expression \( \frac{2x^3 + x^2 + 3}{x+1} \). The degree of the polynomial in the numerator, which is 3, is higher than the degree of the denominator, which is 1. This indicates that the function will tend towards infinity, but the direction (positive or negative infinity) depends on the leading coefficient and the highest degree term's behavior as \( x \to -\infty \).
Polynomial Division
Polynomial division is a technique used to simplify rational functions, especially useful when finding limits at infinity. By dividing each term of a polynomial by the highest power of \( x \) present in the fraction, we can identify the dominant behavior of the function.
This process is akin to long division but applied for polynomials. For the function \( \frac{2x^3 + x^2 + 3}{x+1} \), we divided each term by \( x^3 \).
This helps break down complex expressions, making it possible to separate and then handle terms as their own limits. By focusing primarily on the term with the highest power, other terms become negligible when \( x \to \pm \infty \), simplifying the calculation of the limit.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as \( x \) approaches a point, usually infinity or some value that makes the function undefined. This can often be seen through horizontal, vertical, or slant asymptotes.
In the context of rational functions like \( \frac{2x^3 + x^2 + 3}{x+1} \), the asymptotic behavior will depend on the relationship between the degrees of the numerator and the denominator.
Because the numerator's degree is higher, this function lacks a horizontal asymptote. Instead, it indicates the presence of an oblique or slant asymptote, meaning the function's graph will approach a line as \( x \to \pm \infty \). Hence, understanding these behaviors is essential for sketching the graph of rational functions and predicting their long-term behavior.

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Most popular questions from this chapter

Escape Velocity An object is projected vertically upward fron the earth's surface with an initial velocity \(v_{0}\) of magnitude less than the escape velocity (the velocity that a projectile should have in order to break free of the earth forever). If only the earth's influence is taken into consideration, then the maximum height reached by the rocket is $$ H=\frac{\iota_{0}^{2} R}{2 g R-v_{0}^{2}} $$ where \(R\) is the radius of the earth and \(g\) is the acceleration due to gravity. a. Show that the graph of \(H\) has a vertical asymptote at \(\nu_{0}=\sqrt{2 g R}\), and interpret your result. b. Use the result of part (a) to find the escape velocity. Take the radius of the earth to be \(4000 \mathrm{mi}\left(g=32 \mathrm{ft} / \mathrm{sec}^{2}\right)\). c. Sketch the graph of \(H\) as a function of \(v_{0}\).

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Sketch the graph of a function having the given properties. $$ \begin{array}{l} f(2)=3, f^{\prime}(2)=0, f^{\prime}(x)<0 \text { on }(-\infty, 0) \cup(2, \infty), \\ f^{\prime}(x)>0 \text { on }(0,2), \lim _{x \rightarrow 0^{-}} f(x)=-\infty, \lim _{x \rightarrow 0^{+}} f(x)=-\infty, \\ \lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow \infty} f(x)=1, f^{\prime \prime}(x)<0 \text { on } \\ (-\infty, 0) \cup(0,3), f^{\prime \prime}(x)>0 \text { on }(3, \infty) \end{array} $$

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