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Chapter 4: Problem 43

It follows from the Substitution Rule that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f(1 / x)\) and \(\lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow 0^{-}} f(1 / x)\) Use these formulas to evaluate the limit. $$ \lim _{x \rightarrow \infty} \frac{2 x^{2}+1}{3 x^{2}-5} $$

Short Answer

Expert verified
The limit is \( \frac{2}{3} \).

Step by step solution

01

Identify the Limit Expression

We start with the limit \( \lim _{x \rightarrow \infty} \frac{2 x^{2}+1}{3 x^{2}-5} \). This is the given expression where \( x \) approaches infinity.
02

Apply the Substitution Rule

According to the substitution rule, \( \lim _{x \rightarrow \infty} f(x) = \lim _{x \rightarrow 0^{+}} f(1 / x) \). Let's transform the expression by substituting \( x = 1/t \), hence when \( x \to \infty \), \( t \to 0^{+} \). The expression becomes \( \lim _{t \rightarrow 0^{+}} \frac{2(1/t)^{2} + 1}{3(1/t)^{2} - 5} \).
03

Simplify the Expression

Rewriting the expression with \( t \), we get \( \lim _{t \rightarrow 0^{+}} \frac{2/t^{2} + 1}{3/t^{2} - 5} \). This simplifies to \( \lim _{t \rightarrow 0^{+}} \frac{2 + t^2}{3 - 5t^2} \) by multiplying the numerator and the denominator by \(t^2\).
04

Evaluate the Limit as t approaches 0

Now, substitute \( t = 0^{+} \): \( \frac{2 + 0}{3 - 5 \cdot 0} = \frac{2}{3} \). Since both terms with \( t^{2} \) vanish, the limit becomes \( \frac{2}{3} \).

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Key Concepts

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

Substitution Rule
The substitution rule in calculus is a powerful tool used to evaluate limits that seem complex at first glance. The idea is to substitute a part of the limit's expression with a simpler variable to make calculations easier.
  • In our exercise, we used the substitution rule by transforming the variable \( x \) into \( 1/t \).
  • This allows the limit from \( x \to \infty \) to be transformed into \( t \to 0^+ \).
  • Substituting variables can often simplify fractions or other algebraic expressions, making it easier to evaluate the limit.
Using substitution helps in rewriting the limit in a way that aligns closely with solving simpler limit problems, such as those where approaches zero rather than infinity.
Limits
Limits are a fundamental concept in calculus. They help us understand the behavior of functions as inputs approach a certain value.
  • In this exercise, we are interested in the behavior of a rational function as \( x \) approaches infinity.
  • Limits can describe the value that a function approaches, even if it never actually reaches that value.
  • Finding limits allows for the understanding of trends and potential asymptotic behavior in functions.
Limits are especially useful in defining derivatives and integrals, making them critical in calculus. The ability to transform and solve limits is indispensable in analyzing different mathematical models and real-world phenomena.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as its input grows larger or smaller. For rational functions, this usually involves understanding how the function behaves when approaching infinity.
  • In our example, understanding the asymptotic behavior allows us to predict that as \( x \to \infty \), the function approaches \( \frac{2}{3} \).
  • This prediction is based on the highest degree terms in the numerator and the denominator.
  • When both the numerator and denominator have the same degree, the limit at infinity is the ratio of the leading coefficients, i.e., \( \frac{2}{3} \).
Recognizing these patterns in asymptotic behavior helps mathematicians and scientists predict long-term trends and outcomes which are crucial in fields like economics, physics, and engineering.
Rational Functions
Rational functions are quotients of two polynomials. They have various properties that make them interesting to study and analyze.
  • The function in our example is given by \( \frac{2x^2 + 1}{3x^2 - 5} \), demonstrating a rational function where the coefficients and degrees of the polynomials play a vital role.
  • These functions can be characterized by behavior such as asymptotes, which occur where the denominator equals zero and the function becomes undefined.
  • Rational functions often approach a horizontal asymptote, determined by the leading coefficients if the degrees of the numerator and denominator are the same.
Studying rational functions provides insights into predicting behaviors and solving complex algebraic expressions, as they are prevalent in both pure and applied mathematics.

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Most popular questions from this chapter

Let \(n\) be a positive integer and \(f(x)=x^{n} .\) Show that the graph of \(f\) has at most one inflection point. Determine those values of \(n\) for which the inflection point exists, and find the inflection point.

A wire of length \(L\) is cut into two pieces. One piece is bent to form a square, and the other is bent to form a circle. Determine the minimum possible value for the \(\operatorname{sum} A\) of the areas of the square and the circle. If the wire is actually cut, is there a maximum value of \(A ?\)

In cach of the following, draw the graph of a continuous function \(f\) having the given propertics. a. \(f\) is increasing and its graph is concave upward on \((-\infty, 0)\), and \(f\) is decreasing and its graph is concave downward on \((0, \infty)\). b. \(f\) is decreasing and its graph is concave upward on \((-\infty, 1), f\) is increasing and its graph is concave upward on \((1,2)\), and \(f\) is decreasing and its graph is concave upward on \((2, \infty)\). c. \(f\) is decreasing and its graph is concave upward on \((-\infty, 1), f\) is increasing and its graph is concave upward on \((1,2)\), and \(f\) is increasing and its graph is concave downward on \((2, \infty)\). d. \(f\) is decreasing and its graph is concave downward on \((-\infty, 0), f\) is increasing and its graph is concave downward on \((0,1), f\) is increasing and its graph is concave upward on \((1,5)\), and \(f\) is decreasing and its graph is concave downward on \((5, \infty)\)

Find the horizontal asymptote of the graph of the function. Then sketch the graph of the function. $$ f(x)=\frac{\sqrt{x+2}}{(x+2)^{2}} $$

Suppose the current \(I(t)\) flowing in an electrical circuit at time \(t\) is given by $$ I(t)=\frac{100}{1+t^{2}}+3 \sin \frac{30 t}{\pi} \text { for } t \geq 0 $$ Show that $$ \lim _{t \rightarrow \infty}\left(I(t)-3 \sin \frac{30 t}{\pi}\right)=0 $$ Thus for large values of \(t, I(t)\) is very nearly equal to \(3 \sin (30 t / \pi)\). The expression \(3 \sin (30 t / \pi)\) is called the steady-state current, and the expression \(100 /\left(1+t^{2}\right)\) is the transient current (since it is significant only for small values of \(t\) ).
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