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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
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2장, 문제 2.6.96

Use formal definitions to prove the limit statements in Exercises 93–96.


lim x → −5 (1 / (x + 5)²) = ∞

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To prove that \( \lim_{x \to -5} \frac{1}{(x + 5)^2} = \infty \), we need to use the formal definition of a limit approaching infinity. This means for every positive number \( M \), there exists a \( \delta > 0 \) such that if \( 0 < |x + 5| < \delta \), then \( \frac{1}{(x + 5)^2} > M \).
Start by considering the expression \( \frac{1}{(x + 5)^2} > M \). This inequality can be rewritten as \( (x + 5)^2 < \frac{1}{M} \).
To find \( \delta \), solve the inequality \( |x + 5| < \sqrt{\frac{1}{M}} \). This gives us the condition for \( \delta \) in terms of \( M \).
Set \( \delta = \sqrt{\frac{1}{M}} \). This choice of \( \delta \) ensures that whenever \( 0 < |x + 5| < \delta \), the inequality \( \frac{1}{(x + 5)^2} > M \) holds true.
Conclude that for every \( M > 0 \), there exists a \( \delta = \sqrt{\frac{1}{M}} \) such that if \( 0 < |x + 5| < \delta \), then \( \frac{1}{(x + 5)^2} > M \). Therefore, \( \lim_{x \to -5} \frac{1}{(x + 5)^2} = \infty \) is proven using the formal definition.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limit Definition

The formal definition of a limit states that for a function f(x), the limit as x approaches a value a is L if, for every ε > 0, there exists a δ > 0 such that whenever 0 < |x - a| < δ, it follows that |f(x) - L| < ε. This definition is crucial for proving limit statements rigorously.
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Behavior of Functions Near Asymptotes

Understanding how functions behave near vertical asymptotes is essential for limit proofs. In this case, as x approaches -5, the denominator (x + 5) approaches zero, causing the function (1 / (x + 5)²) to increase without bound, leading to the conclusion that the limit is infinity.
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Epsilon-Delta Proofs

Epsilon-delta proofs are a method used to rigorously establish limits. In this context, one must show that for any large number M (representing infinity), there exists a corresponding δ such that if x is within δ of -5 (but not equal to -5), then (1 / (x + 5)²) exceeds M, confirming the limit statement.
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Left, Right, & Midpoint Riemann Sums