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Continuity

Emily Davis

Emily Davis

5 min read

Next Topic - Non-removable Discontinuities

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Study Guide Overview

This guide covers removable discontinuities, focusing on their definition, identification, and removal. It explains how to determine if a discontinuity is removable by evaluating limits. The guide provides worked examples, practice questions, and exam strategies for handling these types of problems. Key terms include continuous function, limit, and removable discontinuity.

Study Notes on Removable Discontinuity

Table of Contents

  1. Introduction to Removable Discontinuity
  2. Definition and Concept
  3. Identifying Removable Discontinuities
  4. Removing Removable Discontinuities
  5. Worked Example
  6. Practice Questions
  7. Glossary
  8. Summary and Key Takeaways
  9. Exam Strategy

Introduction to Removable Discontinuity

Understanding removable discontinuities is crucial for mastering continuity in functions, a fundamental topic in calculus. This guide will help you identify and remove these discontinuities, ensuring a smooth learning experience.

Definition and Concept

What is a Removable Discontinuity?

A removable discontinuity is a type of discontinuity in a function that can be "removed" to make the function continuous over an interval that includes the discontinuity.

Key Concept

A function fff is continuous at the point x=cx=cx=c if:

- f(c)f(c)f(c) exists, - limโก_xโ†’cf(x)\lim\_{{x \to c}} f(x)lim_xโ†’cf(x) exists, - and limโก_xโ†’cf(x)=f(c)\lim\_{{x \to c}} f(x) = f(c)lim_xโ†’cf(x)=f(c).

Characteristics of Removable Discontinuity

A removable discontinuity is essentially a "hole" in the function. It is a point where:

  • limโกxโ†’cf(x)\lim_{{x \to c}} f(x)limxโ†’cโ€‹f(x) exists,
  • but f(c)f(c)f(c) does not exist.
For instance, consider the function g(x)=xxg(x) = \frac{x}{x}g(x)=xxโ€‹. This function is not continuous at x=0x=0x=0 because g(0)g(0)g(0) does not exist. However, the discontinuity at x=0x=0x=0 is **removable** because limโก_xโ†’0g(x)=1\lim\_{{x \to 0}} g(x) = 1lim_xโ†’0g(x)=1. The limit exists at the point of discontinuity.

Identifying Removable Discontinuities

To identify a removable discontinuity:

  1. Check if limโกxโ†’cf(x)\lim_{{x \to c}} f(x)limxโ†’cโ€‹f(x) exists.
  2. Verify that f(c)f(c)f(c) does not exist or is not equal to the limit.
Common Mistake

Students often confuse removable discontinuities with non-removable ones. Ensure you check both the existence of the limit and the function value at the point.

Removing Removable Discontinuities

How to Remove a Removable Discontinuity

To remove a removable discontinuity, you need to redefine the function at the point of discontinuity:

  1. Define f(c)f(c)f(c) to be equal to limโกxโ†’cf(x)\lim_{{x \to c}} f(x)limxโ†’cโ€‹f(x).
  2. This redefinition makes the function continuous at that point.
For example, redefine the function ggg by: g(x)={xxxโ‰ 01x=0g(x) = \begin{cases} \frac{x}{x} & x \neq 0 \\\\ 1 & x = 0 \end{cases}g(x)=โŽฉโŽจโŽงโ€‹xxโ€‹1โ€‹x๎€ =0x=0โ€‹ This redefinition removes the discontinuity at x=0x=0x=0.

Worked Example

Let fff be the function defined by f(x)=x2+3xxf(x) = \frac{x^2 + 3x}{x}f(x)=xx2+3xโ€‹.

(a) Explain why fff is not continuous at x=0x=0x=0.

Answer:

At x=0x=0x=0: f(0)=02+3โ‹…00=00f(0) = \frac{0^2 + 3 \cdot 0}{0} = \frac{0}{0}f(0)=002+3โ‹…0โ€‹=00โ€‹

Since 00\frac{0}{0}00โ€‹ is not defined, fff is not continuous at x=0x=0x=0.

(b) Explain how the discontinuity at x=0x=0x=0 can be removed.

Answer:

First, find the limit as xxx approaches 0 by factorizing and simplifying: x2+3xx=x(x+3)x=x+3\frac{x^2 + 3x}{x} = \frac{x(x + 3)}{x} = x + 3xx2+3xโ€‹=xx(x+3)โ€‹=x+3

Then, limโกxโ†’0f(x)=0+3=3\lim_{{x \to 0}} f(x) = 0 + 3 = 3xโ†’0limโ€‹f(x)=0+3=3

The limit exists, so the discontinuity at x=0x=0x=0 is removable. Redefine the function so that: f(0)=limโกxโ†’0f(x)=3f(0) = \lim_{{x \to 0}} f(x) = 3f(0)=xโ†’0limโ€‹f(x)=3

Thus, the redefined function is: f(x)={x2+3xxxโ‰ 03x=0f(x) = \begin{cases} \frac{x^2 + 3x}{x} & x \neq 0 \\ 3 & x = 0 \end{cases}f(x)={xx2+3xโ€‹3โ€‹x๎€ =0x=0โ€‹

This removes the discontinuity.

Practice Questions

Practice Question

1. Identify the removable discontinuities in the following functions and redefine them to remove the discontinuities:

(a) h(x)=x2โˆ’1xโˆ’1h(x) = \frac{x^2 - 1}{x - 1}h(x)=xโˆ’1x2โˆ’1โ€‹

(b) k(x)=sinโก(x)xk(x) = \frac{\sin(x)}{x}k(x)=xsin(x)โ€‹

(c) m(x)=x2โˆ’4xโˆ’2m(x) = \frac{x^2 - 4}{x - 2}m(x)=xโˆ’2x2โˆ’4โ€‹

2. Determine if the following function has a removable discontinuity at x=2x=2x=2:

f(x)=x2โˆ’4xโˆ’2f(x) = \frac{x^2 - 4}{x - 2}f(x)=xโˆ’2x2โˆ’4โ€‹

3. Explain why g(x)=x3โˆ’xxg(x) = \frac{x^3 - x}{x}g(x)=xx3โˆ’xโ€‹ is not continuous at x=0x=0x=0 and how to remove the discontinuity.

Glossary

  • Removable Discontinuity: A discontinuity that can be "removed" by redefining the function at the point of discontinuity.
  • Limit: The value that a function approaches as the input approaches some value.
  • Continuous Function: A function without any breaks, jumps, or holes in its graph.

Summary and Key Takeaways

Key Points

  • A removable discontinuity is a "hole" in a function where the limit exists, but the function value does not.
  • To remove a removable discontinuity, redefine the function at the discontinuous point to equal the limit.
  • Understanding limits and continuity is crucial for identifying and removing removable discontinuities.

Key Takeaways

  • Always check the existence of the limit and the function value at the point of discontinuity.
  • Redefine the function to make it continuous at the point of discontinuity.

Exam Strategy

Exam Tip

When facing questions on continuity, always verify the existence of the function value and the limit separately.

Exam Tip

If a function is not continuous at a point, check if the discontinuity is removable by finding the limit.

Exam Tip

Practice identifying and removing removable discontinuities with different types of functions to build confidence.

Good luck with your studies, and remember, practice makes perfect!

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Question 1 of 8

A function fff is continuous at x=cx=cx=c if which of the following conditions are met? ๐Ÿค”

f(c)f(c)f(c) exists, limโกxโ†’cf(x)\lim_{x \to c} f(x)limxโ†’cโ€‹f(x) exists, and limโกxโ†’cf(x)=f(c)\lim_{x \to c} f(x) = f(c)limxโ†’cโ€‹f(x)=f(c)

f(c)f(c)f(c) does not exist, limโกxโ†’cf(x)\lim_{x \to c} f(x)limxโ†’cโ€‹f(x) exists, and limโกxโ†’cf(x)โ‰ f(c)\lim_{x \to c} f(x) \neq f(c)limxโ†’cโ€‹f(x)๎€ =f(c)

limโกxโ†’cf(x)\lim_{x \to c} f(x)limxโ†’cโ€‹f(x) exists, but f(c)f(c)f(c) does not exist

f(c)f(c)f(c) exists, but limโกxโ†’cf(x)\lim_{x \to c} f(x)limxโ†’cโ€‹f(x) does not exist