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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 2, Problem 87

Suppose f(x) = {x^2 − 5x + 6 / x − 3 if x≠3
a if x=3.
Determine a value of the constant a for which lim x→3 f(x) = f(3).

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1
Identify that the problem requires finding the limit of f(x) as x approaches 3 and ensuring it equals f(3).
Recognize that f(x) is a piecewise function, and for x ≠ 3, f(x) = \( \frac{x^2 - 5x + 6}{x - 3} \).
Factor the numerator \( x^2 - 5x + 6 \) to simplify the expression: \( (x - 2)(x - 3) \).
Cancel the common factor \( (x - 3) \) in the numerator and denominator, simplifying f(x) to \( x - 2 \) for x ≠ 3.
Evaluate the limit \( \lim_{{x \to 3}} (x - 2) \) and set it equal to a, ensuring continuity at x = 3.

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Key Concepts

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

Limits

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. In this case, we are interested in the limit of f(x) as x approaches 3. For the limit to exist, the values of f(x) must approach a specific number as x gets closer to 3 from both sides.
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Continuity

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. In this scenario, we need to find a value for the constant 'a' such that the limit of f(x) as x approaches 3 equals f(3), ensuring that the function is continuous at x = 3.
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Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this problem, f(x) is defined differently for x = 3 compared to x ≠ 3. Understanding how to evaluate and manipulate piecewise functions is crucial for determining the appropriate value of 'a' that maintains the function's continuity at x = 3.
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