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Chapter 5: Problem 3

Use the Strategy to prove that the following function is discontinuous at 2 $$ f(x)= \begin{cases}x-\frac{1}{2}, & \text { if } x>2 \\ 1, & \text { if } x=2 \\\ x-1, & \text { if } x<2\end{cases} $$

Short Answer

Expert verified
The function is discontinuous at \(x = 2\) because the limits approaching from the left and right do not match.

Step by step solution

01

Identify the Function Pieces

The function is defined in three parts:- For \(x > 2\), the function is \(f(x) = x - \frac{1}{2}\).- At \(x = 2\), the function is \(f(x) = 1\).- For \(x < 2\), the function is \(f(x) = x - 1\).
02

Define Continuity at a Point

A function is continuous at a point \(x = c\) if the following three conditions are met:1. \(f(c)\) is defined.2. The limit \(\lim_{{x \to c}} f(x)\) exists.3. \(\lim_{{x \to c}} f(x) = f(c)\).
03

Evaluate f(2)

Checking the value of the function at \(x = 2\), we find that \(f(2) = 1\). This satisfies the first condition for continuity.
04

Compute One-Sided Limits as x Approaches 2

Calculate the limit as \(x\) approaches 2 from the right:\(\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} \left(x - \frac{1}{2}\right) = 2 - \frac{1}{2} = 1.5\).Calculate the limit as \(x\) approaches 2 from the left:\(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (x-1) = 2 - 1 = 1\).
05

Determine Limit Existence at x = 2

For the limit \(\lim_{{x \to 2}} f(x)\) to exist, the one-sided limits \(\lim_{{x \to 2^+}} f(x)\) and \(\lim_{{x \to 2^-}} f(x)\) must be equal. Here, \(1.5 eq 1\), so the limit does not exist.
06

Conclude Discontinuity at x = 2

Since the limit as \(x\) approaches 2 does not exist, the function does not satisfy the conditions for continuity at \(x = 2\). Thus, the function is discontinuous at \(x = 2\).

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Key Concepts

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

Piecewise Function
A piecewise function is a function that is defined by different expressions depending on the value of the input. These functions are useful when a relationship behaves differently depending on the interval of the input values. In the exercise, the function is defined as:
  • For \(x > 2\), \(f(x) = x - \frac{1}{2}\).
  • At \(x = 2\), \(f(x) = 1\).
  • For \(x < 2\), \(f(x) = x - 1\).
Each segment of the function handles a specific section of the number line. As a result, piecewise functions can have different limits and continuity properties in each interval.
One-Sided Limits
In calculus, a one-sided limit refers to the value that a function approaches as the input approaches a specific point from one particular direction.
This can either be from the left (denoted as \(x \to c^-\)) or from the right (denoted as \(x \to c^+\)).
For one-sided limits to be equal at a point, the value of the function should smoothly transition from one side to the other.
  • When approaching 2 from the right (\(x \to 2^+\)), the function follows the rule \(f(x) = x - \frac{1}{2}\), so the limit is \(1.5\).
  • When approaching 2 from the left (\(x \to 2^-\)), it follows \(f(x) = x - 1\), making the limit \(1\).
In this instance, the one-sided limits are different which indicates a lack of smooth transition, signifying a discontinuity at \(x = 2\).
Continuity at a Point
Continuity at a point is a key concept in calculus which occurs if a function is unbroken at that point. A function is continuous at \(x = c\) if:
  • The function \(f(c)\) is well-defined.
  • The limit \(\lim_{{x \to c}} f(x)\) exists.
  • The value of the limit equals the function value: \(\lim_{{x \to c}} f(x) = f(c)\).
In the exercise, while \(f(2) = 1\) is well defined, the left and right limits as \(x\) approaches 2 are different. Hence the overall limit \(\lim_{{x \to 2}} f(x)\) does not exist, resulting in a discontinuity at that point.
Limits in Calculus
Limits are fundamental to calculus, providing a way to understand how functions behave near specific points or along specific paths. The concept of a limit explains the tendency of a function as its argument approaches a particular point. It鈥檚 a cornerstone for defining many other calculus concepts including continuity and derivatives.
In the given problem, limits help us determine the properties of the function around \(x = 2\).
Since the one-sided limits are not equal at this point, the overall limit does not exist.
  • This non-existence of a limit at points of the piecewise function signifies that it鈥檚 discontinuous at \(x = 2\).
  • Understanding limits thus allows us to detect and describe discontinuities effectively.
By exploring limits, both from the left and the right of a particular point, we can visualize and interpret the behavior of complex functions better.

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Most popular questions from this chapter

Use the \(\varepsilon-\delta\) definition to prove that \(\lim _{x \rightarrow 1} x^{3}=1\). Hint: \(\quad\) Try \(\delta=\min \left\\{1, \frac{1}{7} \varepsilon\right\\}\).

Write down a function \(f\) defined on the interval \([-1,3]\) for which \(\lim _{x \rightarrow 0^{-}} f(x)\) does not exist but \(\lim _{x \rightarrow 0^{+}} f(x)=1\). Verify that \(f\) has the specified properties.

Determine whether the following limits exist: (a) \(\lim _{x \rightarrow 0} \frac{x^{2}+x}{x}\); (b) \(\lim _{x \rightarrow 0} \frac{|x|}{x}\)

Use the Strategy to prove that \(f(x)=x^{3}\) is continuous at \(1 .\) Hint: Try \(\delta=\min \left\\{1, \frac{1}{7} \varepsilon\right\\} .\)

Let \(a_{n}=\frac{(-1)^{n}}{n^{2}}, n=1,2, \ldots .\) How large must we take \(X\) in order that: (a) \(\left|a_{n}-0\right|<0.1, \quad\) for all \(n>X\) ? (b) \(\left|a_{n}-0\right|<0.01\), for all \(n>X\) ? (c) \(\left|a_{n}-0\right|<\varepsilon, \quad\) for all \(n>X\), where \(\varepsilon\) is a given positive number?
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