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Chapter 2: Problem 28

The statement " \(3 x \rightarrow 7\) as \(x \rightarrow 1^{\prime \prime}\) is false. Choose an \(\varepsilon\) for which no \(\delta\) can be found. The statement \({ }^{4} 3 \mathrm{x} \rightarrow 3\) as \(x \rightarrow 1^{\prime \prime}\) is true. For \(\varepsilon=\frac{1}{2}\) choose a suitable \(\delta\).

Short Answer

Expert verified
No delta exists for \(\varepsilon = 3.5\) (false statement), \(\delta = \frac{1}{6}\) for \(\varepsilon = \frac{1}{2}\) (true statement).

Step by step solution

01

Understanding the Problem

We are given two statements involving limits. One of them is false, and we are asked to find an \(\varepsilon\) for which no \(\delta\) can be found. The other statement is true, and we need to find a suitable \(\delta\) for a given \(\varepsilon\).
02

Analyzing the False Statement

The false statement is \(3x \rightarrow 7\) as \(x \rightarrow 1\). For this statement to be false, no matter how close \(x\) gets to 1, \(3x\) can never equal 7. This is evident because if \(x=1\), then \(3x=3\), not 7. Hence, choose \(\varepsilon = 3.5\) (since \(|3-7| = 4| > 3.5\) and we cannot find any \(x\) such that \(|3x - 7| < 3.5 \) within any neighborhood of 1).
03

Understanding the True Statement

The true statement is \(3x \rightarrow 3\) as \(x \rightarrow 1\). This is a straightforward case: if \(x\) approaches 1, \(3x\) will approach 3.
04

Choosing \(\delta\) for the True Statement

We need to find \(\delta\) given \(\varepsilon = \frac{1}{2}\). We want \(\left|3x - 3\right| < \frac{1}{2}\). This simplifies to \(\left|3(x - 1)\right| < \frac{1}{2}\) which means \(\left|x-1\right| < \frac{1}{6}\). Therefore, the suitable \(\delta\) is \(\frac{1}{6}\).

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Key Concepts

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

Limits
In calculus, limits help us understand the behavior of functions as they approach specific points. A limit is like a target that the outputs of a function get closer to, as the inputs get closer to some value. When we say the limit of a function exists at a point, we mean the function approaches a specific value as the input nears that point. This is a key concept in understanding calculus.
Here are a few important points about limits:
  • Limits help us predict what happens at points where the function might not be directly evaluated, such as points of discontinuity or edge cases.
  • In the exercise, the statement "3x approaches 7 as x approaches 1" is exploring the limit of the function 3x as x nears the value 1.
  • The statement's truthfulness or falsehood is based on whether the outputs of the function, 3x, do indeed get closer to 7 as x gets closer to 1.
A strong grasp of limits will lay the foundation for more advanced calculus topics, like derivatives and integrals. Think of limits as zooming in closer and closer to a point to see how a function behaves.
Epsilon-delta definition
The epsilon-delta definition is a rigorous way to prove that a function approaches a limit. It might seem complex at first, but it can be quite intuitive. The idea is simple: for every small number \(\epsilon\) (epsilon), there is a small "neighborhood" around our point of interest, \(\delta\) (delta), such that as long as we're within this \(\delta\) neighborhood, the function's values remain within the \(\epsilon\) of the limit.
Let's break it down:
  • \(\epsilon\) represents how close we want the function's value to be to the limit. It is a positive number chosen arbitrarily small.
  • \(\delta\) is the distance we are willing to go away from the point of interest. It is also a positive number, and it depends on \(\epsilon\).
  • If for every \(\epsilon > 0\), there is a \(\delta > 0\) such that whenever \(|x - a| < \delta\), \(|f(x) - L| < \epsilon\), then \(\lim_{x \to a} f(x) = L\) is true.
In the exercise, we considered different values of \(\epsilon\) and \(\delta\) to show why a given statement is false, and to prove why a limit statement holds true.
Problem-solving
Problem-solving in calculus often involves using theoretical concepts to analyze the behavior of functions. When faced with statements about limits, we need to examine whether the statements are true or false and provide justifications for our conclusions.
A typical approach might include:
  • Identifying known quantities and what needs to be proven or disproven.
  • Applying the epsilon-delta definition to validate or invalidate the limit statements.
  • Using algebraic manipulation to express inequalities and reach conclusions.
In the given exercise, we analyzed two statements about limits. One was true, and the other was false. By choosing specific epsilon values, we observed whether a suitable delta could be found. If not, the statement was demonstrated to be false, illustrating a practical application of these concepts.
Continuity
Continuity of functions is another essential aspect in calculus closely tied to limits. Intuitively, a function is continuous at a point if there is no abrupt change in its value鈥攁 "smooth" transition. Mathematically, a function \(f(x)\) is continuous at a point \(x = c\) if:
1. \(f(c)\) is defined.
2. \(\lim_{x \to c} f(x) = f(c)\).
3. \(\lim_{x \to c} f(x) = f(c)\).
This ensures that the limit as we approach the point \(c\) equals the function's actual value at that point.
Key highlights include:
  • Continuity implies that small changes in input (x) result in small changes in the output (f(x)).
  • For small \(\delta\) values close to \(c\), the values of \(f(x)\) should be close to \(f(c)\).
  • In the exercise, continuity was considered when analyzing limit statements, determining where changes occur smoothly and consistently.
Understanding continuity helps us grasp not just limits but also how functions behave across intervals, laying the groundwork for differential and integral calculus.

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Most popular questions from this chapter

The sum of \(1+r+r^{2}+\cdots+r^{n-1}\) is \(a_{n}=\left(1-r^{n}\right) /(1-r)\). What is the limit of \(a_{n}\) as \(n \rightarrow \infty\) ? For which \(r\) does the limit exist?

Name two functions with \(d f / d x=1 / x^{2}\).

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 0} \frac{2 x \tan x}{\sin x} $$

Find the distances \(f(t)\), starting from \(f(0)=0\), to match these velocities: (a) \(v(t)=\cos t \sin t\) (b) \(v(t)=\tan t \sec ^{2} t\) (c) \(v(t)=\sqrt{1+t}\)

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 5} \frac{x^{2}+25}{x-5} $$
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