Continuity (Epsilon-Delta)

**ah-bee** · May 1st 2008, 05:33 AM

How would i approach a question that asks me to use the Cauchy Definition (Epsilon-Delta) to prove that f(x) = x is continuous at x = 1.

**flyingsquirrel** · May 1st 2008, 06:09 AM

Hi

What is wanted : $\forall \varepsilon>0,\ \exists \ \delta \ | \ |x-1|<\delta \Rightarrow |f(x)-1|=|x-1|<\varepsilon$ .

The aim is to show that there exists such a $\delta$ for each $\varepsilon$ . ( $\delta$ will depend on $\varepsilon$ but can't depend on $x$ ). It can be achieved by taking $\varepsilon>0$ , and looking for a condition on $|x-1|$ so that $|f(x)-1|<\varepsilon$ .

An example :

Let's show the continuity of $x \mapsto x^2$ at $0$ .

What is wanted : $\forall \varepsilon>0,\ \exists \ \delta \ | \ |x-0|<\delta \Rightarrow |x^2-0|<\varepsilon$ which can be rewritten $\forall \varepsilon>0,\ \exists \ \delta \ | \ |x|<\delta \Rightarrow x^2<\varepsilon \ (1)$

What condition has to be respected by $x$ so that $x^2<\varepsilon$ ? Take the square root of this inequality ( $x\mapsto\sqrt{x}$ being an increasing function) : $x^2<\varepsilon \Leftrightarrow x<\sqrt{\varepsilon}$ and here is the condition. Hence, by taking $\delta=\sqrt{\varepsilon}$ , (1) is true and we've shown the continuity of the function at 0.

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