/*! 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 57 $$\text { Let } f(x)=\left\\{\be... [FREE SOLUTION] | 91Ó°ÊÓ

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$$\text { Let } f(x)=\left\\{\begin{array}{ll}x, & x<1 \\\x+1, & x>1\end{array}\right.$$ FUGURES CABT COPY a. Let \(\epsilon=1 / 2 .\) Show that no possible \(\delta>0\) satisfies the following condition: For all \(x, \quad 0<|x-1|<\delta \Rightarrow|f(x)-2|<1 / 2\) That is, for each \(\delta>0\) show that there is a value of \(x\) such that \(0<|x-1|<\delta \quad\) and \(\quad|f(x)-2| \geq 1 / 2\) This will show that \(\lim _{x \rightarrow 1} f(x) \neq 2\) b. Show that \(\lim _{x \rightarrow 1} f(x) \neq 1\) c. Show that \(\lim _{x \rightarrow 1} f(x) \neq 1.5\)

Short Answer

Expert verified
No limit exists for \(f(x)\) as \(x\to1\).

Step by step solution

01

Understanding f(x) Near x=1

The piecewise function given is \(f(x) = \left\{ \begin{array}{ll} x, & x<1 \ x+1, & x>1\end{array}\right.\). We need to examine the behavior of \(f(x)\) as \(x\) approaches 1 from both sides to determine if any specified limit conditions are satisfied.
02

Analyzing the Limit for f(x) as x Approaches 1 for ε=1/2

Given \(\epsilon = 1/2\), we look for \(\delta > 0\) such that if \(0 < |x-1| < \delta\), then \(|f(x)-2| < 1/2\). For \(x < 1\), \(f(x) = x\) and \(|f(x) - 2| = |x - 2|\) which is always greater than 1/2 as \(x < 1\). For \(x > 1\), \(f(x) = x + 1\) and \(|f(x) - 2| = |x + 1 - 2| = |x - 1|\) which is equal to or greater than 1/2 when \(x > 1.5\). Thus, no \(\delta\) satisfies the condition where \(|f(x)-2| < 1/2\), confirming \(\lim_{x \to 1} f(x) eq 2\).
03

Analyzing the Limit for f(x) as x Approaches 1 with Respect to 1

To show \(\lim_{x \to 1} f(x) eq 1\), consider \(f(x)\) for \(x < 1\) and \(x > 1\). For \(x < 1,\) \(f(x) = x\) and as \(x\) approaches 1, \(f(x)\) approaches 1. For \(x > 1\), \(f(x) = x + 1\) and \(f(x)\) approaches 2 as \(x\) approaches 1. Because the limit from the left and right do not match, \(\lim_{x \to 1} f(x) eq 1\).
04

Verifying the Limit Does Not Exist for 1.5

Consider the behavior of \(f(x)\) around \(x = 1\) for \(\lim_{x \to 1} f(x) = 1.5\). For \(x < 1\), \(f(x) = x\) approaches 1 as \(x\to 1^-.\) For \(x > 1\), \(f(x) = x+1\) approaches 2 as \(x\to 1^+.\) Given these differing behaviors, \(|f(x) - 1.5|\) does not consistently meet an \(\epsilon\)-condition (\(\epsilon = 0.5\)) near \(x=1\), confirming \(\lim_{x \to 1} f(x) eq 1.5\).
05

Conclusion

\(f(x)\) does not approach any single value as \(x\) approaches 1 from both sides. Consequently, \(\lim_{x \to 1} f(x)\) does not exist.

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Key Concepts

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

Piecewise Functions
In calculus, piecewise functions are defined by different expressions based on different intervals of the input variable. These functions can exhibit different behaviors across specified domains. For example, our function is defined as:
- For values of \(x\) less than 1, \(f(x) = x\)- For values of \(x\) greater than 1, \(f(x) = x + 1\)
Piecewise functions can create situations where the behavior of a function changes sharply at certain points, like at \(x = 1\) in this example. This can make finding limits tricky, as you need to consider the behavior of the function from both sides of the point of interest.
Understanding piecewise functions is essential since they often appear in real-world scenarios where a situation might change suddenly, like pricing tiers in economics or step functions in control processes.
Epsilon-Delta Definition
The epsilon-delta definition is a formal way to define the limit of a function at a particular point. It's crucial for proving rigorously whether a limit exists. According to this definition, a limit \(L\) exists at a point \(a\) if, for every \(\epsilon > 0\) (no matter how small!), there exists a \(\delta > 0\) such that if the distance from \(x\) to \(a\) is less than \(\delta\) but not zero (\(0 < |x-a| < \delta\)), then the distance from \(f(x)\) to \(L\) is less than \(\epsilon\) (\(|f(x) - L| < \epsilon\)).
This definition ensures that we can trap \(f(x)\) closely around \(L\) whenever \(x\) is sufficiently close to \(a\). It's a rigorous way to handle the intuitive idea of a limit. When using piecewise functions, the epsilon-delta approach can help decide the existence of a limit when direct substitution isn't feasible, as it was in this case where the function had discontinuous behavior at \(x = 1\).
Limit Does Not Exist
A limit does not exist when a function's output does not approach a single value as the input approaches a particular point from both directions. In the case of our function, as \(x\) approaches 1, the function behaves differently based on whether \(x\) approaches 1 from the left or the right:
- From the left, \(x < 1\), \(f(x) = x\), and \(f(x)\) approaches 1.- From the right, \(x > 1\), \(f(x) = x + 1\), and \(f(x)\) approaches 2.
Here, the limit does not exist at \(x = 1\) because the values approached from the left and right do not match. This is a typical scenario where a limit does not exist, and confirms the piecewise nature of our function. Such understanding is vital in calculus to correctly interpret points of discontinuity.

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Most popular questions from this chapter

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