Question

continuity

Answer 1

How do i show that \[f(x)=x^{2}\] is continous using the epsilon-delta definition?

Answer 2

\[f:(\mathbb{R} ,d) \rightarrow (\mathbb{R},d)\] is defined by \[f(x) = x ^{2}\] with \[x \epsilon X\] and d the usual metric

Answer 3

\[|x-a|<\delta, \text{need \to show that } |f(x)-f(a)|<\epsilon\\ |f(x)-f(a)|=|x^2-a^2|=|(x-a)(x+a)|=|x-a||x+a|\\=|x-a||x-a+a+a|\\|x-a||x-a+2a|\le|x-a|(|x-a|+2a)<\delta(\delta+2a)=\epsilon\\\delta=\frac{-2a\mp\sqrt{4a^2+4*1*\epsilon}}{2}=-a\mp\sqrt{a^2+\epsilon}\\\ \text{since} \,\delta >0, \,\delta=-a+\sqrt{a^2+\epsilon}\\\text{\let }\epsilon>0 \text{ be given. Choose }\delta=-a+\sqrt{a^2+\epsilon}\,.\\\text{Then |f(x)-f(a)|<}\epsilon\,,\text{whenever}\,|x-a|<\delta\]

Answer 4

thank u a lot.