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 forany $l$ 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 $0<|u-x_{0}|<\\delta$ andput $l=f(v)$ then substituting in the condition with $u=x_{l}$ we get $|f(x_{l})-l|<\\epsilon$ . A contradiction.