proof of Cauchy condition for limit of function

The forward direction is . Assume that $\lim_{x\to x_{0}}f(x)=L$ . Then given $\epsilon$ there is a $\delta$ such that

|f(u)-L|<\epsilon/2\text{ when }0<|u-x_{0}|<\delta.

Now for $0<|u-x_{0}|<\delta$ and $0<|v-x_{0}|<\delta$ we have

|f(u)-L|<\epsilon/2\text{ and }|f(v)-L|<\epsilon/2

and so

|f(u)-f(v)|=|f(u)-L-(f(v)-L)|\leq|f(u)-L|+|f(v)-L|<\epsilon/2+\epsilon/2=\epsilon.

We prove the reverse by contradiction. Assume that the condition holds. Now suppose that $\lim_{x\to x_{0}}f(x)$ does not exist. This means that for any $l$ and any $\epsilon$ sufficiently small then for any $\delta>0$ there is $x_{l}$ such that $0<|xl-x_{0}|<\delta~{}\text{and}~{}|f(x_{l})-l|\geq\epsilon$ . For any such $\epsilon$ choose $u$ such that $0<|u-x_{0}|<\delta$ and put $l=f(v)$ then substituting in the condition with $u=x_{l}$ we get $|f(x_{l})-l|<\epsilon$ . 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