Math 9C Sec 2.3

Math 9C Sec 2.3 - ˃ ˃ ˂ ˂ If a limit doesn’t exists...

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2.2 Recall that = limx afx L if for all Epsilon is greater than 0 there exists a delta 0 a such that |x-a| delta ˃ ˂ implies |f(x)-L| Epsilon. ˂ Definition → - = limx a fx L if for all Epsilon is greater than 0, there exists a delta greater than 0 such that is a-delta is less than x which is less that a then |f(x)-L| Epsilon. ˂ Ex) Prove → + = limx 0 x 0 a x a+delta ˂ ˂ |f(x)-L| Epsilon ˂ 0 x delta ˂ ˂ | - x 0 | Epsilon ˂ | x | Epsilon ˂ -Epsilon ˂x Epsilon ˂ 0 ˂x Epsilon ˂ 0 x Eps squared ˂ ˂ Let Epsilon 0 be given. Let delta=Epsilon squared. Assume 0 x delta(=Eps squared). Then, |f(x)-L=| ˃ ˂ ˂ x - 0|=| x |= x˂delta (=Eps squared) If a limit exists, for all Epsilon 0, there’s a delta 0 such that |x-a| delta implies |f(x)-L| Epsilon.
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Unformatted text preview: ˃ ˃ ˂ ˂ If a limit doesn’t exists, for all Epsilon 0, such that for all delta 0, |x-a| delta, but |f(x)-L|≥ Epsilon. ˃ ˃ ˂ Ex) Let f(x)= 0 if x 0 and 1 if x≥0 ˂ Prove → limx 0fx doesn’t exist. Let Epsilon=1/2. Assume → = limx 0fx L |x-a| Delta ˂ |f(x)-L| Epsilon ˂ |x| delta ˂ |f(x)-L| 1/2 ˂-delta x 0 ˂ ˂ 0≤x 8 ˂ |f(x)-L| 1/2 ˂ |f(x)-L| 1/2 ˂ |0-L| 1/2 ˂ |1-L| 1/2 ˂-1/2 L 1/2 ˂ ˂-1/2 1-L 1/2 ˂ ˂ 3/2 L 1/2 ˃ ˃ Contradiction between the two....
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  • Spring '07
  • APOORVA
  • Math, Existence, |x-a|˂delta implies |f, |x-0|˂Epsilon |x|˂Epsilon -Epsilon˂x˂Epsilon, |x-a|˂ delta, limx→0fx=L |x-a|˂Delta |x|˂delta

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Math 9C Sec 2.3 - ˃ ˃ ˂ ˂ If a limit doesn’t exists...

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