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Answer by Monadologie for If $|x-x_0|
Everything is correct except the last sentence. You can say:The result is clear since$$|(x+y)-(x_0+y_0)| \le |x-x_0|+|y-y_0|< \epsilon.$$
View ArticleAnswer by azif00 for If $|x-x_0|
It is correct, but you could have written it as follows :Suppose $|x-x_0|<\varepsilon /2$ and $|y-y_0|<\varepsilon /2$. Now$$\begin{align}|(x+y)-(x_0+y_0)| &= |(x-x_0)+(y-y_0)| \\&\leq...
View ArticleIf $|x-x_0|
This is a problem from Spivak's Calculus 4th ed., Chapter 1If $|x-x_0|<\frac {\varepsilon}{2}$ and $|y-y_0|<\frac {\varepsilon}{2}$ then $|(x+y)-(x_0+y_0)|<\varepsilon$ and...
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