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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 2.7.47

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


limx→0(1x2+1)=∞{\(\displaystyle\)\(\lim\)_{x\(\to\)0}}\(\left\)(\(\frac{1}{x^2}\)+1\(\right\))=\(\infty\)

Guida verificata passo dopo passo
1
Step 1: Understand the definition of an infinite limit. The limit of a function f(x) as x approaches a value c is infinity if for every positive number M, there exists a δ > 0 such that if 0 < |x - c| < δ, then f(x) > M.
Step 2: Identify the function and the point of interest. Here, the function is f(x) = \( \frac{1}{x^2} + 1 \) and we are interested in the behavior as x approaches 0.
Step 3: Set up the inequality based on the definition. We need to show that for every M > 0, there exists a δ > 0 such that if 0 < |x| < δ, then \( \frac{1}{x^2} + 1 > M \).
Step 4: Simplify the inequality \( \frac{1}{x^2} + 1 > M \) to \( \frac{1}{x^2} > M - 1 \). This implies \( x^2 < \frac{1}{M - 1} \) and thus |x| < \( \frac{1}{\sqrt{M - 1}} \).
Step 5: Choose δ = \( \frac{1}{\sqrt{M - 1}} \). This choice of δ ensures that whenever 0 < |x| < δ, the inequality \( \frac{1}{x^2} + 1 > M \) holds, proving the limit is infinity.

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Infinite Limits

Infinite limits occur when the value of a function increases without bound as the input approaches a certain point. In this context, we analyze the behavior of the function as x approaches 0. If the function's value grows larger and larger, we denote this behavior as approaching infinity, which is a key aspect of understanding limits in calculus.
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