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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 2.2.57
Chapter 2, Problem 2.2.57

Limits of Average Rates of Change


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

Verified step by step guidance
1
First, understand that the expression limh→0 (f(x+h) − f(x)) / h represents the derivative of the function f(x) at a specific point x. This is the definition of the derivative using the limit process.
Identify the function f(x) = x² and the point x = 1 where you need to evaluate the limit. This means you are finding the derivative of f(x) = x² at x = 1.
Substitute f(x) = x² into the limit expression: limh→0 ((1+h)² − 1²) / h. This involves substituting x = 1 into the function and considering the increment h.
Expand the expression (1+h)² to get 1 + 2h + h². This is a crucial step to simplify the expression inside the limit.
Simplify the expression: ((1 + 2h + h²) − 1) / h, which reduces to (2h + h²) / h. Then, factor out h from the numerator to get h(2 + h) / h, and cancel h from the numerator and denominator, leaving 2 + h. Finally, evaluate the limit as h approaches 0, which gives the derivative at x = 1.

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Key Concepts

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

Limits

In calculus, a limit describes the value that a function approaches as the input approaches a certain point. It is fundamental for understanding continuity, derivatives, and integrals. The notation lim h→0 indicates that we are examining the behavior of a function as the variable h approaches zero, which is crucial for defining instantaneous rates of change.
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One-Sided Limits

Derivative

The derivative of a function at a point represents the instantaneous rate of change of the function with respect to its variable. It is defined as the limit of the average rate of change (the slope of the secant line) as the interval approaches zero. For the function f(x) = x², the derivative at x = 1 can be found using the limit definition, yielding the slope of the tangent line at that point.
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Derivatives

Secant and Tangent Lines

A secant line intersects a curve at two or more points, providing an average rate of change between those points. In contrast, a tangent line touches the curve at a single point and represents the instantaneous rate of change at that point. The relationship between secant and tangent lines is essential in calculus, as the limit of the secant line's slope as the two points converge gives the derivative.
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Related Practice
Textbook Question

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


limx →4 (9 − x) = 5

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Textbook Question

Limits as x → ∞ or x → −∞


The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x. Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36. Write ∞ or −∞ where appropriate.


lim x→∞ (x − 3) / √(4x² + 25)

348
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Textbook Question

Limits of Rational Functions


In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.


f(x) = (x + 1)/(x² + 3)

365
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Textbook Question

Finding Limits of Differences When x → ±∞


Find the limits in Exercises 84–90. (Hint: Try multiplying and dividing by the conjugate.)


lim x → ∞ (√(x + 9) − √(x + 4))

325
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Textbook Question

Use the Intermediate Value Theorem in Exercises 69–74 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.

x³ − 15x + 1 = 0 (three roots)

338
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Textbook Question

Limits with trigonometric functions


Find the limits in Exercises 43–50.


limx→−π √(x + 4) cos(x + π)

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