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 ProofOfCauchyConditionForLimitOfFunction 2013-03-22 18:59:08 2013-03-22 18:59:08 puff (4175) puff (4175) 8 puff (4175) Proof msc 54E35 msc 26A06 msc 26B12