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Chapter 9: Problem 13

Find the limit. $$ \text { 3. } \lim _{k \rightarrow \infty} \frac{k+1}{k} $$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Recognize the type of limit

You need to evaluate the limit \( \lim_{k \rightarrow \infty} \frac{k+1}{k} \). This is a rational function where the degrees of the numerator and the denominator are the same. Thus, we can simplify it to find the limit as \( k \to \infty \).
02

Factor and simplify

Divide both the numerator and the denominator by the highest power of \( k \), which is \( k \) itself in this case. This gives us \( \frac{k+1}{k} = \frac{k}{k} + \frac{1}{k} = 1 + \frac{1}{k} \).
03

Evaluate the limit

Now find \( \lim_{k \rightarrow \infty} (1 + \frac{1}{k}) \). As \( k \) approaches infinity, \( \frac{1}{k} \) approaches 0. Thus, \( 1 + \frac{1}{k} \) approaches 1.

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Key Concepts

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

Rational Functions
A rational function is a type of function that is represented by the ratio of two polynomials. In mathematical terms, a rational function is defined as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). Rational functions appear frequently in calculus because they are used to model various real-world scenarios.
  • Rational functions can have asymptotes, which are lines that the graph of the function approaches but never touches.
  • The simplest rational function is \( \frac{1}{x} \), which serves as a base model for understanding more complex behaviors.
  • When analyzing limits involving rational functions as variables approach infinity, the degrees of the numerator and the denominator are an important consideration.
In the exercise above, the given function \( \frac{k+1}{k} \) is a rational function. The numerator and denominator have equal degrees, both being linear polynomials.
Simplifying Expressions
Simplifying expressions is a crucial step in solving calculus problems, especially when dealing with rational functions. The goal of simplifying is to rewrite the expression in a form that is easier to work with or evaluate.
To simplify \( \frac{k+1}{k} \), we divide both the numerator and denominator by the highest power of \( k \) present, which in this case is \( k \) itself.
  • Start with \( \frac{k+1}{k} \).
  • Divide each term by \( k \): \( \frac{k}{k} + \frac{1}{k} = 1 + \frac{1}{k} \).
  • Because \( \frac{k}{k} = 1 \), this simplifies the expression.
By reducing the original expression to \( 1 + \frac{1}{k} \), it becomes much more straightforward to evaluate as \( k \) approaches infinity.
Evaluating Infinite Limits
Evaluating infinite limits involves understanding the behavior of a function as the variable within it grows larger and larger, essentially heading towards infinity. Infinite limits can illustrate how functions behave in the long run.
For the expression \( \lim_{k \rightarrow \infty} (1 + \frac{1}{k}) \), as \( k \) grows infinitely large, \( \frac{1}{k} \) approaches zero. This is because dividing one by larger and larger numbers results in a value tending towards zero.
Thus, \( 1 + \frac{1}{k} \) approaches \( 1 \), and the limit is \( 1 \).
  • Analyzing the behavior of each term in the expression as \( k \rightarrow \infty \) is crucial.
  • Identifying which terms tend to disappear (approaching zero) can simplify your calculation.
  • Recognizing that small additions or subtractions from a constant term diminish over infinite expansions helps in finding accurate limits.
Infinite limits reveal how expressions simplify over vast scales, providing insights into their foundational properties and behaviors.

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Most popular questions from this chapter

A pendulum \(L\) meters long swings back and forth, with maximum angle of deflection \(\phi\). Let \(x=\sin \phi / 2\), and let \(g=32\) (feet per second per second) be the acceleration due to gravity. Then the period of oscillation \(T\) of the pendulum is given by the formula The Foucault Pendulum in the Smithsonian Institution in Washington, D.C. (Smithsonian Institution) $$ T(x)=2 \pi \sqrt{\frac{L}{g}}\left(1+\sum_{n=1}^{\infty} \frac{1^{2} \cdot 3^{2} \cdot 5^{2} \cdots(2 n-1)^{2}}{2^{2} \cdot 4^{2} \cdot 6^{2} \cdots(2 n)^{2}} x^{2 n}\right) $$ The series on the right converges for \(|x|<1\) (see Exercise 23). Notice that \(\left|\frac{T(x)}{2 \pi \sqrt{L / g}}-1\right|=\sum_{n=1}^{\infty} \frac{1^{2} \cdot 3^{2} \cdot 5^{2} \cdots(2 n-1)^{2}}{2^{2} \cdot 4^{2} \cdot 6^{2} \cdots(2 n)^{2}} x^{2 n}\) Usually \(\phi\) is very small in pendulum clocks, say, \(0<\phi \leq\) \(\pi / 12\), and consequently $$ 0

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