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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 2, Problem 2.3.40

Using the Formal Definition


Prove the limit statements in Exercises 37–50.


lim x→0 √(4 − x) = 2

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Start by recalling the formal definition of a limit: For the limit \( \lim_{x \to a} f(x) = L \), for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \).
In this problem, we need to prove that \( \lim_{x \to 0} \sqrt{4 - x} = 2 \). This means for every \( \epsilon > 0 \), we need to find a \( \delta > 0 \) such that if \( 0 < |x - 0| < \delta \), then \( |\sqrt{4 - x} - 2| < \epsilon \).
Consider the expression \( |\sqrt{4 - x} - 2| < \epsilon \). To manipulate this, multiply and divide by the conjugate: \( |\sqrt{4 - x} - 2| = \frac{|(\sqrt{4 - x} - 2)(\sqrt{4 - x} + 2)|}{|\sqrt{4 - x} + 2|} = \frac{|4 - x - 4|}{|\sqrt{4 - x} + 2|} = \frac{|x|}{|\sqrt{4 - x} + 2|} \).
To ensure \( \frac{|x|}{|\sqrt{4 - x} + 2|} < \epsilon \), we need \( |x| < \epsilon \cdot |\sqrt{4 - x} + 2| \). Since \( \sqrt{4 - x} \) is close to 2 when \( x \) is near 0, \( |\sqrt{4 - x} + 2| \) is close to 4. We can choose \( \delta \) such that \( |x| < \epsilon \cdot 4 \).
Finally, choose \( \delta = \min(1, 4\epsilon) \) to ensure that \( |x| < \delta \) implies \( |\sqrt{4 - x} - 2| < \epsilon \). This completes the proof using the formal definition of a limit.

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Key Concepts

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

Limit Definition

The formal definition of a limit, often called the epsilon-delta definition, states that for a function f(x) to have a limit L as x approaches a value c, for every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε. This definition is crucial for rigorously proving limit statements.
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Square Root Function

The square root function, √(x), is a fundamental mathematical function that returns the principal square root of a non-negative number x. Understanding its behavior, especially near points of interest like x = 0, is essential for analyzing limits involving square roots, as it affects the continuity and differentiability of the function.
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Limit Properties

Limit properties, such as the limit of a sum, product, or composition of functions, help simplify complex limit problems. For instance, knowing that the limit of a constant is the constant itself, and the limit of a function as x approaches a point can be distributed over addition and multiplication, aids in breaking down and solving limit problems efficiently.
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