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Let $$ f(x)=\sum_{k=0}^{n} a_{k} x^{k} \text { and } g(x)=\sum_{k=0}^{m} b_{k} x^{k}\left(a_{n} \neq 0, b_{m} \neq 0\right) $$ Find \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}\) if (i) \(n>m ;\) (ii) \(n

Short Answer

Expert verified
(i) \(\infty\), (ii) 0, (iii) \(\frac{a_n}{b_m}\).

Step by step solution

01

Understanding the Problem

Given two polynomial functions, \( f(x) \) and \( g(x) \), where \( f(x) = \sum_{k=0}^{n} a_{k} x^{k} \) and \( g(x) = \sum_{k=0}^{m} b_{k} x^{k} \), we are tasked with finding the limit of their quotient as \( x \to \infty \). Polynomials are often compared using their leading terms, especially for behavior at infinity.
02

Consider When Degree of f(x) is Greater than g(x)

If \( n > m \), the degree of the numerator \( f(x) \) is higher than that of the denominator \( g(x) \). As \( x \to \infty \), the highest power in \( f(x) \), which is \( a_n x^n \), will dominate. The same logic applies to \( g(x) \) with term \( b_m x^m \). This implies, \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{a_n x^n}{b_m x^m} = \lim_{x \to \infty} \frac{a_n}{b_m} x^{n-m} = \infty \). Thus, the limit is infinity.
03

Consider When Degree of f(x) is Less than g(x)

If \( n < m \), the degree of the numerator is lower than that of the denominator. As \( x \to \infty \), the leading term in \( g(x) \) will grow faster than any term in \( f(x) \). Therefore, \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{a_n x^n}{b_m x^m} = \lim_{x \to \infty} \frac{a_n}{b_m x^{m-n}} = 0 \). Thus, the limit is 0.
04

Consider When Degrees of f(x) and g(x) are Equal

If \( n = m \), the dominant terms in both numerator and denominator are \( a_n x^n \) and \( b_m x^m \) respectively, resulting in \( \frac{a_n x^n}{b_m x^m} = \frac{a_n}{b_m} \). In this case, \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{a_n}{b_m} \). Therefore, the limit is the ratio of the leading coefficients \( \frac{a_n}{b_m} \).

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Key Concepts

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

Polynomial Degree
To understand polynomials deeply, a key concept to grasp is the polynomial degree. This simply refers to the highest power of the variable, typically denoted as \( x \), in the polynomial. The degree is an important characteristic that influences how the polynomial behaves, especially when taking limits. For example, in the function \( f(x) = 3x^3 + 4x^2 + 2x + 7 \), the degree is 3, because the highest exponent is 3. The term that contains the highest exponent is called the leading term, and it is immensely powerful in determining the polynomial’s behavior, especially as \( x \) approaches infinity.
  • The degree of a polynomial is the highest power of \( x \) with a non-zero coefficient.
  • It plays a critical role in evaluating limits at infinity.
  • The leading term becomes more significant as \( x \) grows larger.
Understanding the degree helps us focus on the leading term of the polynomial, which is crucial for predicting the function's behavior at extreme values of \(x\).
Leading Term Comparison
When you compare polynomial functions, the leading term is your ultimate guide. The leading term is the part of a polynomial where \( x \) is raised to the greatest power. When taking the limit of a quotient of two polynomials, the leading terms in the numerator and the denominator usually determine the result. Consider two scenarios:1. **If the degree of the numerator is greater than the degree of the denominator:** The leading term of the numerator will grow faster as \( x \) approaches infinity. Consequently, the fraction approaches infinity as well.2. **If the degree is equal in numerator and denominator:** Both grow at the same pace. Here, the limit resolves to the ratio of the leading coefficients \( \frac{a_n}{b_m} \). By focusing solely on the leading terms, we simplify complex polynomials to a comparative fraction of their respective leading terms as \( x \to \infty \). This method simplifies our calculations immensely.
Behavior at Infinity
The behavior of polynomial functions as \( x \) approaches infinity is predictable once you understand their leading terms and degrees. Polynomials demonstrate consistent growth patterns at infinity, which allows us to infer limits easily. Here's what happens in different cases:
  • Higher degree in the numerator: If \( n > m \), the polynomial function in the numerator grows faster. Thus, the quotient approaches infinity.
  • Higher degree in the denominator: If \( n < m \), growth is slower in the numerator. The quotient shrinks to zero as \( x \) increases.
  • Equal degrees: If \( n = m \), both functions grow at the same rate, making the limit the ratio of the leading coefficients.
This systematic approach allows us to leverage algebraic properties for quick estimations of limits and behaviors, illustrating that in many cases, only the highest degree terms matter.

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Most popular questions from this chapter

Prove that if the functions \(f_{n}\) and \(g_{n}\) are real or complex (or if the \(g_{n}\) are vector valued and the \(f_{n}\) are scalar valued), and if $$ f_{n} \rightarrow f \text { and } g_{n} \rightarrow g \text { (uniformly) on } B \text { , } $$ then $$ f_{n} g_{n} \rightarrow f g \text { (uniformly) on } B $$ provided that either \(f\) and \(g\) or the \(f_{n}\) and \(g_{n}\) are bounded on \(B\) (at least from some \(n\) onward \() ;\) cf. Problem 11 . Disprove it for the case where only one of \(f\) and \(g\) is bounded.

Prove that if \(f\) is monotone on \((a, b) \subseteq E^{*},\) it has at most countably many discontinuities in \((a, b)\).

Prove that every compact set is complete. Disprove the converse by examples.

Prove Theorem 2 , with (i) replaced by the weaker assumption ("subuniform limit") $$ (\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in G_{\neg p}(\delta)\right)\left(\forall y \in G_{\neg q}(\delta)\right) \quad \rho(g(x), f(x, y))<\varepsilon $$ and with iterated limits defined by $$ s=\lim _{x \rightarrow p} \lim _{y \rightarrow q} f(x, y) $$ iff \((\forall \varepsilon>0)\) $$ \left(\exists \delta^{\prime}>0\right)\left(\forall x \in G_{\neg p}\left(\delta^{\prime}\right)\right)\left(\exists \delta_{x}^{\prime \prime}>0\right)\left(\forall y \in G_{\neg q}\left(\delta_{x}^{\prime \prime}\right)\right) \quad \rho(f(x, y), s)<\varepsilon $$

A set is said to be totally disconnected iff its only connected subsets are one-point sets and \(\emptyset\). Show that \(R\) (the rationals) has this property in \(E^{1}\).

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