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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.7.42
Chapter 2, Problem 2.7.42

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.
lim x→5 1/x^2=1/25

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Start by recalling the precise definition of a limit: For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 5| < \delta \), then \( \left| \frac{1}{x^2} - \frac{1}{25} \right| < \varepsilon \).
Simplify the expression \( \left| \frac{1}{x^2} - \frac{1}{25} \right| \) to find a relationship between \( \varepsilon \) and \( \delta \). This can be rewritten as \( \left| \frac{25 - x^2}{25x^2} \right| \).
Factor the numerator \( 25 - x^2 \) as \( (5 - x)(5 + x) \), so the expression becomes \( \left| \frac{(5 - x)(5 + x)}{25x^2} \right| \).
To ensure \( \left| \frac{(5 - x)(5 + x)}{25x^2} \right| < \varepsilon \), we need to bound \( |5 + x| \) and \( |x^2| \) by choosing \( \delta \) such that \( |x - 5| < \delta \) implies \( |x - 5| < 1 \), which gives \( 4 < x < 6 \).
Within this interval, \( |5 + x| \) is bounded by 11, and \( x^2 \) is bounded below by 16, so choose \( \delta \) such that \( \delta = \min\left(1, \frac{16\varepsilon}{11}\right) \) to satisfy the condition \( \left| \frac{(5 - x)(5 + x)}{25x^2} \right| < \varepsilon \).

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Key Concepts

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

Limit Definition

The precise definition of a limit states that for a function f(x) to approach a limit L as x approaches a value a, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε. This formalism is essential for rigorously proving limits in calculus.
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Epsilon-Delta Relationship

In the context of limits, the ε (epsilon) represents how close f(x) must be to the limit L, while δ (delta) represents how close x must be to the point a. Establishing a relationship between ε and δ is crucial for demonstrating that as x gets sufficiently close to a, f(x) will be within ε of L, thus proving the limit exists.
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Continuous Functions

A function is continuous at a point if the limit as x approaches that point equals the function's value at that point. In this case, since 1/x^2 is continuous around x = 5, we can directly apply the limit definition without concerns about discontinuities, simplifying the proof process.
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Related Practice
Textbook Question

Evaluate each limit and justify your answer. 

lim x→4 √x^3−2x^2−8x / x−4

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Textbook Question

Use the precise definition of infinite limits to prove the following limits.


limx→−11(x+1)4=∞{\(\displaystyle\]\lim\)_{x\(\to\)-1}}\(\frac{1}{\left(x+1\right)^4}\)=\(\infty\)

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Textbook Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→1 (8x+5)=13

Textbook Question

Determine limx→∞f(x)\(\lim\)_{x\(\rightarrow\]\infty\)}f\(\left\)(x\(\right\)) and limx→−∞f(x)\(\lim\)_{x\(\rightarrow\)-\(\infty\)}f\(\left\)(x\(\right\)) for the following functions. Then give the horizontal asymptotes of ff (if any).


f(x)=12x4−4x8−9x4f\(\left\)(x\(\right\))=\(\frac{1}{2x^4-\sqrt{4x^8-9x^4}\)}

Textbook Question

Use the precise definition of a limit to prove the following limits. Specify a relationship between ε and δ that guarantees the limit exists.

lim x→4 x^2−16 / x−4=8 (Hint: Factor and simplify.)

Textbook Question

Determine the following limits at infinity.


lim x→−∞ 3x^11