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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 2.2.58
Chapter 2, Problem 2.2.58

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 = -2

Verified step by step guidance
1
Identify the function f(x) = x² and the point x = -2 where we need to evaluate the limit.
Substitute f(x) = x² into the limit expression: lim(h→0) [(f(x+h) - f(x)) / h]. This becomes lim(h→0) [((x+h)² - x²) / h].
Expand the expression (x+h)² to get x² + 2xh + h². Substitute this back into the limit expression: lim(h→0) [(x² + 2xh + h² - x²) / h].
Simplify the expression by canceling out x²: lim(h→0) [(2xh + h²) / h].
Factor out h from the numerator: lim(h→0) [h(2x + h) / h]. Cancel h from the numerator and denominator, resulting in lim(h→0) [2x + h]. Now, substitute x = -2 and evaluate the limit as h approaches 0.

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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 behavior of a function as its input approaches a certain value. It is fundamental for understanding continuity, derivatives, and integrals. The notation lim h→0 (f(x+h) - f(x)) / h specifically represents the limit of the average rate of change of a function as the interval approaches zero, which leads to the concept of the derivative.
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One-Sided Limits

Derivatives

The derivative of a function at a point quantifies 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 as the interval shrinks to zero. In the context of the given question, finding the derivative of f(x) = x² at x = -2 involves evaluating the limit expression provided.
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Secant and Tangent Lines

Secant lines connect two points on a curve and represent the average rate of change between those points. As the two points get infinitely close, the secant line approaches the tangent line, which touches the curve at a single point and represents the instantaneous rate of change. Understanding this relationship is crucial for grasping how limits lead to derivatives in calculus.
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Related Practice
Textbook Question

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


lim x→0 x sin (1/x) = 0


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Textbook Question

Domains and Asymptotes


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


y = 2x / (x² − 1)

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Textbook Question

If functions f(x) and g(x) are continuous for 0 ≤ x ≤ 1, could f(x)/g(x) possibly be discontinuous at a point of [0,1]? Give reasons for your answer.

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Textbook Question

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


limx→3 (3x − 7) = 2

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Textbook Question

Infinite Limits


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


lim x→0 4 / x²/⁵

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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→∞ (2√x + x⁻¹) / (3x − 7)

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