proof of Cauchy condition for limit of function
The forward direction is . Assume that limx→x0f(x)=L. Then given ϵ there is a δ such that
|f(u)-L|<ϵ/2 when 0<|u-x0|<δ. |
Now for 0<|u-x0|<δ and 0<|v-x0|<δ we have
|f(u)-L|<ϵ/2 and |f(v)-L|<ϵ/2 |
and so
|f(u)-f(v)|=|f(u)-L-(f(v)-L)|≤|f(u)-L|+|f(v)-L|<ϵ/2+ϵ/2=ϵ. |
We prove the reverse by contradiction.
Assume that the condition holds.
Now suppose that limx→x0f(x) does not exist. This means that for
any l
and any ϵ sufficiently small then for any δ>0 there is
xl such that 0<|xl-x0|<δand|f(xl)-l|≥ϵ.
For any such ϵ choose u such that 0<|u-x0|<δ and
put l=f(v) then substituting in the condition with u=xl we get
|f(xl)-l|<ϵ. A contradiction.
Title | proof of Cauchy condition for limit of function |
---|---|
Canonical name | ProofOfCauchyConditionForLimitOfFunction |
Date of creation | 2013-03-22 18:59:08 |
Last modified on | 2013-03-22 18:59:08 |
Owner | puff (4175) |
Last modified by | puff (4175) |
Numerical id | 8 |
Author | puff (4175) |
Entry type | Proof |
Classification | msc 54E35 |
Classification | msc 26A06 |
Classification | msc 26B12 |