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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 2.7.45

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


limx→41(x−4)2=∞{\(\displaystyle\)\(\lim\)_{x\(\to\)4}}\(\frac{1}{\left(x-4\right)^2}\)=\(\infty\)

Guida verificata passo dopo passo
1
Step 1: Understand the definition of an infinite limit. The statement \( \lim_{{x \to 4}} \frac{1}{{(x-4)^2}} = \infty \) means that for every positive number \( M \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 4| < \delta \), then \( \frac{1}{{(x-4)^2}} > M \).
Step 2: Start by manipulating the inequality \( \frac{1}{{(x-4)^2}} > M \). This can be rewritten as \( (x-4)^2 < \frac{1}{M} \).
Step 3: Solve the inequality \( (x-4)^2 < \frac{1}{M} \) for \( x \). This gives \( |x-4| < \frac{1}{\sqrt{M}} \).
Step 4: Choose \( \delta = \frac{1}{\sqrt{M}} \). This choice of \( \delta \) ensures that whenever \( 0 < |x - 4| < \delta \), the inequality \( \frac{1}{{(x-4)^2}} > M \) holds true.
Step 5: Conclude that since for every \( M > 0 \), there exists a \( \delta = \frac{1}{\sqrt{M}} \) such that \( 0 < |x - 4| < \delta \) implies \( \frac{1}{{(x-4)^2}} > M \), the limit \( \lim_{{x \to 4}} \frac{1}{{(x-4)^2}} = \infty \) is proven.

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