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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 2.5.30
Chapter 2, Problem 2.5.30

At what points are the functions in Exercises 13–30 continuous?
f(x) = { (x³ − 8)/(x² − 4), x ≠ 2, x ≠ −2
3, x = 2
4, x = −2

Verified step by step guidance
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Step 1: Understand the definition of continuity. A function is continuous at a point if the limit of the function as it approaches the point from both sides is equal to the function's value at that point.
Step 2: Identify the points where the function might not be continuous. These are typically points where the function is not defined or where the expression changes form. In this case, the function is defined differently at x = 2 and x = -2.
Step 3: Check the continuity at x = 2. Calculate the limit of the function as x approaches 2 from both sides. Use the expression \( \frac{x^3 - 8}{x^2 - 4} \) for x ≠ 2 and compare it to the value of the function at x = 2, which is 3.
Step 4: Check the continuity at x = -2. Calculate the limit of the function as x approaches -2 from both sides using the expression \( \frac{x^3 - 8}{x^2 - 4} \) for x ≠ -2 and compare it to the value of the function at x = -2, which is 4.
Step 5: Determine the intervals where the function is continuous. The function is continuous at all points except where the limits do not match the function values at x = 2 and x = -2. Verify the continuity by ensuring the limits match the function values at these points.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Continuity of Functions

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. This means there are no breaks, jumps, or holes in the graph of the function at that point. For a function to be continuous everywhere, it must be continuous at every point in its domain.
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Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. To determine continuity, we often evaluate the limit of the function at points where it may not be defined, such as where the denominator is zero. If the limit exists and equals the function's value, the function is continuous at that point.
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Piecewise Functions

Piecewise functions are defined by different expressions based on the input value. In this case, the function f(x) has different definitions for x values of 2 and -2 compared to other values. Understanding how to evaluate these functions at specific points is crucial for determining their continuity and ensuring that the function behaves as expected across its domain.
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Related Practice
Textbook Question

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Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.


y = 4 + 3x² / (x² + 1)

Textbook Question

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Find the limits in Exercises 37–48. Write ∞ or −∞ where appropriate.


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Prove the limit statements in Exercises 37–50.


limx→1 f(x) = 1 if f(x) = {x², x ≠ 1

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In Exercises 3–8, find the limit of each function (a) as x → ∞ and (b) as x → −∞. (You may wish to visualize your answer with a graphing calculator or computer.)


f(x) = 2/x − 3

Textbook Question

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Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh→0 (f(x+h) − f(x)) / h occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function f.


f(x) = x², x = 1

1
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Textbook Question

Limits with trigonometric functions


Find the limits in Exercises 43–50.


limx→−π √(x + 4) cos(x + π)