Proof that f(x) = 1/x is Continuous on (0, infinity) using Delta-Epsilon

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Proof that f(x) = 1/x is Continuous on (0, infinity) using Delta-Epsilon
  1. The Math Sorcerer
  2. advanced calculus
  3. mathematics
  4. math
  5. delta epsilon proofs
  6. proof
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Комментарии
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Wow this series is a goldmine for my analysis course, thank you so much!

tristanthepterodactyl
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at 06:43 The reverse triangle inequality is |x - y| >= ||x| - |y|| . how does |x_0 + (x - x_0)| >= |x_0| - |x - x_0| match that pattern?

ElizaberthUndEugen
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I believe a proof should consider both values of delta
separately dependent on their relative values.

The scratchwork seems to use the two delta values
simultaneously and it does not need to!

TheCliveAC
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How is it proved for any x in the interval if you have only proved the continuity for the arbitrarily chosen interval xo +/- xo/2? why didn't you choose a different interval?

seansethi
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4:18 is it okay to just bound it abs(x-x0)/xx0 < abs(x-x0)/1? since denominator for 1 is bigger than any number of xx0?

gjsksn
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How does this change if you are proving continuity from (-infinity, 0). Obviously the part of choosing the delta is going to change, but it would then affect the min case... as x_0/2 will always be less than x_0^2 is x_0<0.

erik
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I have seen epsilon delta proofs for other functions too, but I haven't understood the reason behind taking minimum ...?

comkunal
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Why does delta need to be the minimum of the two values that we found?

joshuapereira
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Why do you use one value for delta and then a different one later in the proof?

lmatton
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Thanks . Now i know why delta is minimum of those two.

amarparajuli
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Hey I have a question please: how could this video show uniform continuity of 1/x for (0, infinity), ( which includes (0, 1) ), yet in another video you say 1/x is not uniformly continuous for (0, 1). I’m so confused!

MathCuriousity
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This video is really helpful. Thank you so much!

蔺美云
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Is there a difference between prooving uniform continuity and continuity on an interval? ( they mean the same thing right?)

melodieeluafi
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Why exactly did you choose x0/2 and not x0/4? Is there any specific reason for that?

bushranaheed
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Is it ok if I choose another interval? Let's say that I want to choose the interval (x_0, 3x_0) which is centered at 2x_0, and then I find 1/x to be smaller then 1/x_0, therefore I found delta to be any real number less than epsilon*x_0*x_0? Is this ok?

raducumihaicristian
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What does delta and epsilon mean/define in limit sense

zzgiantdwarf
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Why do you have to choose between two deltas? I don't understand why you defined the first delta, the second is just fine to me.

MrCentrax
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hi, how come you chose both of the different deltas to be the minimum in the different equations in the proof?

kapi
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Thank you; this proof is not intuitive at first glance.

TN-pjlk
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I think there is a better proof for the claim that lim 1/x (x —> c) = 1/c for all 0 < c. If 0 < δ & 0 < x & -δ < x – c < δ, then c – δ < x < c + δ. Hence, if you can find δ such that δ < c, then 0 < c – δ, and 1/x < 1/(c – δ, thus 1/(c·x) = 1/|c·x| < 1/[c·(c – δ)], and |x – c|/|c·x| = |1/x – 1/c| < |x – c|/[c·(c – δ)] < δ/[c·(c – δ)]. Therefore, |1/x – 1/c| < δ/[c·(c – δ)]. To prove |1/x – 1/c| < ε, let ε = δ/[c·(c – δ)], equivalent to δ = c^2·ε/(1 + c·ε). Thus, all that remains to be proven is that c^2·ε/(1 + c·ε) < c, in accordance to δ < c, and the proof is complete. c^2·ε/(1 + c·ε) < c is equivalent to c·ε/(1 + c·ε) < 1, equivalent to c·ε < 1 + cε, equivalent to 0 < 1, which is axiomatic. Q. E. D.

angelmendez-rivera