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Chapter 2: Problem 23

Find the largest \(\delta\) that "works" for the given \(\epsilon\) $$\lim _{x \rightarrow 1} 2 x=2 ; \epsilon=0.1$$

Short Answer

Expert verified
The largest working \(\delta\) for the given \(\epsilon = 0.1\) is \(\delta = 0.05\).

Step by step solution

01

Setup a \( \delta - \epsilon \) proof

Considering the given limit \( \lim_{x \rightarrow 1} 2x = 2 \), start by stating that \(0 < |x-1| < \delta \), and it needs to be shown that \( |f(x) - L| < \epsilon \). Here, \(f(x) = 2x \) and \(L = 2\). This simplifies the latter statement to \( |2x - 2| < \epsilon \).
02

Simplify

The statement \( |2x - 2| < \epsilon \) can further be simplified to \( |2(x - 1)| < \epsilon \). This means our \( \delta \) is potentially \( \frac{\epsilon}{2} \).
03

Finding the largest \( \delta \)

Given \(\epsilon = 0.1\), if we substitute it into the simplified \( \delta \), we find \( \delta = \frac{\epsilon}{2} = \frac{0.1}{2} = 0.05 \). Thus, \( \delta = 0.05 \) is a 'working' \(\delta\) for \(\epsilon = 0.1\). However, this is the smallest working \( \delta \). We are asked to find the largest. For \(\epsilon = 0.1\), any \( \delta \leq 0.05\) would work. Therefore, the largest working \(\delta\) in this context is 0.05.

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Key Concepts

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

Limit of a Function
In calculus, the concept of the limit of a function is pivotal. It helps us understand the behavior of functions as they approach certain points. Specifically, when we say the limit of a function as x approaches a value is L, such as \( \lim_{x \to 1} 2x = 2 \), it implies that as x gets closer and closer to 1, the function \( 2x \) gets closer to 2.
The limit does not concern whether x actually reaches the value 1. Instead, it is about the trend. The function's output gets indefinitely close to the limit as x approaches the given point.
  • This idea is crucial for ensuring continuity in functions.
  • Limits form the foundation for defining more complex concepts like derivatives and integrals.
Choosing Delta
To rigorously define what it means for a function to approach a limit, we introduce the concept of \( \delta \). In our scenario, for the limit \( \lim_{x \to 1} 2x = 2 \) with a tolerance \( \epsilon = 0.1 \), we need to find a \( \delta \) such that when x is within \( \delta \) distance from 1, the function's value is within \( \epsilon \) distance from 2.
Here's a simplified approach:
  • Rewrite the condition \( |2x - 2| < \epsilon \) as \( |2(x - 1)| < \epsilon \).
  • Simplify it to find\( |x - 1| < \frac{\epsilon}{2} \), which gives \( \delta = \frac{\epsilon}{2} \).
This ensures the function remains within the desired range of the limit value. The choice of \( \delta \) guarantees the validity of the limit under a given \( \epsilon \).
Epsilon-Delta Proof
An \( \epsilon-\delta \) proof provides a formal way of proving limits. Consider our example: We need to prove that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) ensuring the function behaves as expected near the point of interest.
For \( \epsilon = 0.1 \) in \( \lim_{x \to 1} 2x \), start by determining a \( \delta \) so that if \( 0 < |x - 1| < \delta \), then \( |2x - 2| < 0.1 \).
Steps involved include:
  • Start with the inequality \( |2(x - 1)| < \epsilon \).
  • Solve for \( \delta \) as \( \delta = \frac{\epsilon}{2} \), yielding \( \delta = 0.05 \) for our \( \epsilon = 0.1 \) example.
This method clarifies that as long as x stays within a distance of \( \delta \) from 1, the function will remain within \( \epsilon \) of its limit. This framework is a cornerstone of mathematical analysis, providing confidence in the rigor of calculus.

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Most popular questions from this chapter

Set \(f(x)=\left\\{\begin{aligned} 1-c x+d x^{2}, & x \leq-1 \\ x^{2}+x, &-1 < x < 2 \text { Find values of } \\ c x^{2}+d x+4, & x \geq 2 \end{aligned}\right.\) \(c\) and \(d\) that make \(f\) continuous on \((-\infty, \infty) .\) Lxe a graphing utility to verify your result.

Use a graphing utility to find at least one number \(c\) at which $$\lim _{x \rightarrow 6} f(x)$$ does not exist. $$f(x)=\frac{x+1}{\left|x^{3}+1\right|}$$

The function \(f\) is not defined at \(x=0\). Use a graphing utility to graph \(f\). Zoom in to determine whether there is a number \(k\) such that the function $$F(x) \quad\left\\{\begin{aligned} f(x), & x \neq 0 \\ k, & x=0 \end{aligned}\right.$$ is continuous at \(x=0\). If so, what is \(k ?\) Support your conclusion by calculating the limit using a CAS. $$f(x)=\frac{x^{2}}{1-\cos 2 x}$$.

Evaluate the limit, taking \(a\) and \(b\) as nonzero constants. $$\lim _{x \rightarrow 0} \frac{\cos a x}{\cos b x}$$

Show that \(\lim _{x \rightarrow 1}|x-1| \sin x=0\)
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