Prove that limx→∞1x=0, with these steps: (a) What is the ε–N statement that must be shown to prove that limx→∞1x=0? (b) Argue that x ∈ (N,∞) if and only if x > N. (c) Argue that 1x ∈ (0−ε, 0+ε) if and only if&...

(a) For all ε > 0, there exists N > 0 such that if x ∈ (N,∞), then 1x ∈ (0 − ε, 0 + ε). (b) x ∈ (N,...

Q. 71 - Calculus | StudyQuestionHub

Q. 71

Question

$Prove that \lim_{x \to \infty} \frac{1}{x} = 0, with these steps : (a) What is the ε - N statement that must be shown to prove that \lim_{x \to \infty} \frac{1}{x} = 0 ? (b) Argue that x \in (N, \infty) if and only if x > N . (c) Argue that \frac{1}{x} \in (0 - ε, 0 + ε) if and only if - ε < \frac{1}{x} < ε . Then argue that for this limit, it suffices to consider 0 < \frac{1}{x} < ε . (d) Given any particular ε, what value of N > 0 would guarantee that if x > N, then 0 < \frac{1}{x} < ε ? Your answer will depend on ε . (e) Put the previous four parts together to prove the limit statement .$

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Answer

$(a) For all ε > 0, there exists N > 0 such that if x \in (N, \infty), then \frac{1}{x} \in (0 - ε, 0 + ε) . (b) x \in (N, \infty) is equivalent to x > N . (c) \frac{1}{x} \in (- ε, ε) means that - ε < \frac{1}{x} < ε . For the limit we are considering, \frac{1}{x} approaches y = 0 from above, so, we actually wish to show that we can guarantee 0 < \frac{1}{x} < ε . (d) For a given ε > 0, choose N = \frac{1}{ε} . (e) Given any ε > 0, let N = \frac{1}{ε} . If x \in (N, \infty) then x > N > 0 by part (b) . Thus by our choice from part (d), 0 < \frac{1}{ε} < x, and thus 0 < \frac{1}{x} < ε . By part (c) this means - ε < \frac{1}{x} < ε and thus \frac{1}{x} \in (0 - ε, 0 + ε) .$

1Part (a) Step 1. The ε - N statement

$For all ε > 0, there exists N > 0 such that if x \in (N, \infty), then \frac{1}{x} \in (0 - ε, 0 + ε) .$

2Part (b) Step 1. Argument regarding N

$Now, x \in (N, \infty) is equivalent to x > N .$

3Part (c) Step 1. Argument regarding ε

$Again, \frac{1}{x} \in (- ε, ε) means that - ε < \frac{1}{x} < ε . For the limit we are considering, \frac{1}{x} approaches y = 0 from above, so, we actually wish to show that we can guarantee 0 < \frac{1}{x} < ε .$

4Part (d) Step 1. Relationship between ε - N

$For a given ε > 0, choose N = \frac{1}{ε} .$

5Part (e) Step 1. Proof of the limit statement

$Given any ε > 0, let N = \frac{1}{ε} . If x \in (N, \infty) then x > N > 0 by part (b) . Thus by our choice from part (d), 0 < \frac{1}{ε} < x, and thus 0 < \frac{1}{x} < ε . By part (c) this means - ε < \frac{1}{x} < ε and thus \frac{1}{x} \in (0 - ε, 0 + ε) .$

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