Math 9C Sec 2.3

Math 9C Sec 2.3 - If a limit doesnt exists, for all Epsilon...

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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 doesnt 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 x0 Prove limx 0fx doesnt 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 0x 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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Math 9C Sec 2.3 - If a limit doesnt exists, for all Epsilon...

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