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.
Hi
What is wanted : $\displaystyle \forall \varepsilon>0,\ \exists \ \delta \ | \ |x-1|<\delta \Rightarrow |f(x)-1|=|x-1|<\varepsilon$.
The aim is to show that there exists such a $\displaystyle \delta$ for each $\displaystyle \varepsilon$. ($\displaystyle \delta$ will depend on $\displaystyle \varepsilon$ but can't depend on $\displaystyle x$). It can be achieved by taking $\displaystyle \varepsilon>0$, and looking for a condition on $\displaystyle |x-1|$ so that $\displaystyle |f(x)-1|<\varepsilon$.
An example :
Let's show the continuity of $\displaystyle x \mapsto x^2$ at $\displaystyle 0$.
What is wanted : $\displaystyle \forall \varepsilon>0,\ \exists \ \delta \ | \ |x-0|<\delta \Rightarrow |x^2-0|<\varepsilon$ which can be rewritten $\displaystyle \forall \varepsilon>0,\ \exists \ \delta \ | \ |x|<\delta \Rightarrow x^2<\varepsilon \ (1)$
What condition has to be respected by $\displaystyle x$ so that $\displaystyle x^2<\varepsilon$ ? Take the square root of this inequality ($\displaystyle x\mapsto\sqrt{x}$ being an increasing function) : $\displaystyle x^2<\varepsilon \Leftrightarrow x<\sqrt{\varepsilon}$ and here is the condition. Hence, by taking $\displaystyle \delta=\sqrt{\varepsilon}$, (1) is true and we've shown the continuity of the function at 0.