/*! 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 41 Suppose that \(f\) is bounded on... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 41

Suppose that \(f\) is bounded on an open interval containing \(x_{0}\). Show that \(\lim _{x \rightarrow x_{0}} f(x)\) exists if and only if $$ \lim _{x \rightarrow x_{0}-} f(x)=\lim _{x \rightarrow x_{0}+} f(x)=\varliminf_{x \rightarrow x_{0}-} f(x)=\lim _{x \rightarrow x_{0}+} f(x) . $$ in which case \(\lim _{x \rightarrow \infty 0} f(x)\) is the common value of these four expressions.

Short Answer

Expert verified
Question: Show that the limit of a function exists at a given point if and only if the left and right limits exist and are equal. Answer: To prove this, we've demonstrated that if the limit exists, then the left and right limits must exist and be equal. Conversely, if the left and right limits exist and are equal, then the overall limit exists. Therefore, the limit of a function exists at a given point if and only if the left and right limits exist and are equal.

Step by step solution

01

Part 1: If \(\lim_{x \rightarrow x_0} f(x)\) exists, then \(\lim_{x \rightarrow x_0^-} f(x)\) and \(\lim_{x \rightarrow x_0^+} f(x)\) exist and are equal.

We are given that \(\lim_{x \rightarrow x_0} f(x) = L\). By definition, for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - x_0| < \delta\), then \(|f(x) - L| < \epsilon\). Let's consider the left and right limits separately. For the left limit, we look at values of \(x\) that are close to \(x_0\) but are to the left of it (i.e., with \(x < x_0\)). If \(0 < x_0 - x < \delta\), then \(|x - x_0| < \delta\). By the definition of the limit, we have \(|f(x) - L| < \epsilon\). Thus, we've shown that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < x_0 - x < \delta\), then \(|f(x) - L| < \epsilon\). Therefore, \(\lim_{x \rightarrow x_0^-} f(x) = L\). For the right limit, we look at values of \(x\) that are close to \(x_0\) but are to the right of it (i.e., with \(x > x_0\)). If \(0 < x - x_0 < \delta\), then \(|x - x_0| < \delta\). By the definition of the limit, we have \(|f(x) - L| < \epsilon\). Thus, we've shown that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < x - x_0 < \delta\), then \(|f(x) - L| < \epsilon\). Therefore, \(\lim_{x \rightarrow x_0^+} f(x) = L\). Since both left and right limits exist and are equal to \(L\), then \(\lim_{x \rightarrow x_0^-} f(x) = \lim_{x \rightarrow x_0^+} f(x) = L\).
02

Part 2: If \(\lim_{x \rightarrow x_0^-} f(x)\) and \(\lim_{x \rightarrow x_0^+} f(x)\) exist and are equal, then \(\lim_{x \rightarrow x_0} f(x)\) exists.

We are given that \(\lim_{x \rightarrow x_0^-} f(x) = \lim_{x \rightarrow x_0^+} f(x) = L\). By definition, for every \(\epsilon > 0\), there exist \(\delta_1, \delta_2 > 0\) such that: 1. If \(0 < x_0 - x < \delta_1\), then \(|f(x) - L| < \epsilon\). 2. If \(0 < x - x_0 < \delta_2\), then \(|f(x) - L| < \epsilon\). Now, let \(\delta = \min\{\delta_1, \delta_2\}\). If \(0 < |x - x_0| < \delta\), then either \(0 < x_0 - x < \delta_1\) or \(0 < x - x_0 < \delta_2\). In either case, we have \(|f(x) - L| < \epsilon\). Thus, we've shown that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - x_0| < \delta\), then \(|f(x) - L| < \epsilon\). Therefore, \(\lim_{x \rightarrow x_0} f(x)\) exists and is equal to \(L\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Limits
In real analysis, the concept of limits is fundamental. Limits help us understand the behavior of a function as the input approaches a particular point. Simply put, if we say \(\lim_{x \rightarrow x_0} f(x) = L\), it means that as \(x\) gets closer to \(x_0\), the function \(f(x)\) gets closer to the value \(L\).

It is crucial to remember that a limit describes the trend of the function, not necessarily the actual value at that point. Therefore, it鈥檚 possible for a function to have a limit at a point even if it is not defined exactly at that point.

The process of finding limits involves understanding both the left-hand and right-hand limits:
  • Left-hand limit \((\lim_{x \rightarrow x_0^-} f(x))\): Examines values of \(x\) approaching \(x_0\) from the left.
  • Right-hand limit \((\lim_{x \rightarrow x_0^+} f(x))\): Considers values of \(x\) as they approach \(x_0\) from the right.
For a limit \(\lim_{x \rightarrow x_0} f(x)\) to exist, both the left-hand and right-hand limits must exist and be equal. This highlights the importance of checking both directions when evaluating limits.
Epsilon-Delta Definition
The epsilon-delta definition is a formal way to define the limit of a function in calculus. It captures the essence of what it means for a function to approach a particular value.

Here's a step-by-step breakdown:
  • We start by considering the statement \(\lim_{x \rightarrow x_0} f(x) = L\).
  • For every small positive number \(\epsilon\) (epsilon represents how close you're willing to be to \(L\)), there exists a small positive number \(\delta\) such that if \(0 < |x - x_0| < \delta\), then \(|f(x) - L| < \epsilon\).
  • This definition implies that we can make \(f(x)\) as close as we want to \(L\) (within \(\epsilon\)) by choosing \(x\) sufficiently close to \(x_0\) (within \(\delta\)).
Essentially, the epsilon-delta definition is about controlling the outputs of a function by restricting its inputs.
Understanding Bounded Functions
A function is said to be bounded on an interval if there is a limit to its values within that interval 鈥� it cannot go to infinity or negative infinity within that space. For any bounded function, you can find a real number such that the function's output never exceeds this number.

Boundedness is an important property because it prevents the function from having excessive growth in value. It makes proving or refuting the existence of limits more feasible, as bounded functions are more predictable across an interval.

In the context of the exercise, knowing that \(f\) is bounded ensures that examining its behavior around \(x_0\) will not be thwarted by unbounded growth or decrease, which helps us apply our limit and epsilon-delta principles effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let $$ f(x)=\left\\{\begin{array}{ll} e^{-1 / x^{2}}, & x \neq 0 \\ 0, & x=0 \end{array}\right. $$ Show that \(f\) has derivatives of all orders on \((-\infty, \infty)\) and every Taylor polynomial of \(f\) about 0 is identically zero. Hivr: See Exercise \(2.4 .40 .\)

Prove: If $$ T_{n}(x)=\sum_{r=0}^{n} \frac{x^{r}}{r !} $$ then $$ T_{n}(x)

Let \(n\) be a positive integer. A function \(f\) has a zero of multiplicity \(n\) at \(x_{0}\) if \(f\) is \(n\) times differentiable on a neighborhood of \(x_{0} . f\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)=\cdots=\) \(f^{(n-1)}\left(x_{0}\right)=0\) and \(f^{(n)}\left(x_{0}\right) \neq 0 .\) Prove that \(f\) has a zero of multiplicity \(n\) at \(x_{0}\) if and only if $$ f(x)=g(x)\left(x-x_{0}\right)^{n} $$ where \(g\) is continuous at \(x_{0}\) and \(n\) times differentiable on a deleted neighborhood of \(x_{0}, g\left(x_{0}\right) \neq 0,\) and $$ \lim _{x \rightarrow x_{0}}\left(x-x_{0}\right)^{j} g^{(j)}(x)=0, \quad 1 \leq j \leq n-1 $$

(a) Let \(F_{n+} G_{n},\) and \(H_{n}\) be the \(n\) th Taylor polynomials about \(x_{0}\) of \(f, g,\) and their product \(h=f g\). Show that \(H_{n}\) can be obtained by multiplying \(F_{n}\) by \(G_{n}\) and retaining only the powers of \(x-x_{0}\) through the \(n\) th. HINT: Use Exerise \(2.5 .8(b) .\) (b) Use the method suggested by (a) to compute \(h^{(r)}\left(x_{0}\right), r=1,2,3,4\). (i) \(h(x)=e^{x} \sin x, \quad x_{0}=0\) (ii) \(h(x)=(\cos \pi x / 2)(\log x), \quad x_{0}=1\) (iii) \(h(x)=x^{2} \cos x, \quad x_{0}=\pi / 2\) (iv) \(h(x)=(1+x)^{-1} e^{-x}, \quad x_{0}=0\)

Suppose that \(f\) is bounded on an interval \(\left(x_{0}, b\right] .\) Show that \(\underline{\varliminf m}_{r \rightarrow x_{0}+} f(x)=\) \(\overline{\lim }_{x \rightarrow x_{0}+} f(x)\) if and only if \(\lim _{x \rightarrow x_{0}+} f(x)\) exists, in which case $$ \lim _{x \rightarrow x_{0}+} f(x)=\lim _{x \rightarrow x_{0}+} f(x)=\lim _{x \rightarrow x_{0}+} f(x) $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.