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- Thread starter brydustin
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mathwonk

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what are you reading? Isn't this explained in every book on several variable calculus? i like courant, volume 2, and wendell fleming's book. basically you define the length of a vector |(a,b,c)| = sqrt(a^2+b^2+c^2), and do all the epsilon delta in those terms, word for word the same as in one variable.

It's differentiability where you have to do something new, because you can't divide vectors.

|x+y| ≤ |x| + |y|, implies continuity of addition.

It's differentiability where you have to do something new, because you can't divide vectors.

|x+y| ≤ |x| + |y|, implies continuity of addition.

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Of course, even with functions of one variable, nobody would use the epsilon-delta kind of argument for an "actual" function something like

[tex]f(x) = \frac{e^{\sin(x)}\sqrt{1+x^3}}{3x^2+7}[/tex]

You prove continuity for the basic functions and use the theorems about products, quotients, compositions etc. The same thing is true for functions of 2 variables.

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Note that continuity along lines parallel to the z and t axes does not imply continuity of the function; considerbut I just don't know how to prove (for example) that f(x,y) = x + y or f(t,z) = t*z is continuous, other than saying something like: Well because f(t)=t is continuous and ...., therefore the composition of....

[tex]

f(z,t) = \left\{\begin{array}{ll}

\frac{|t|e^{-\frac{|t|}{z^2}}}{z^2} & z\neq 0 \\

0 & z=0 \end{array}

[/tex]

in the neighborhood of (0,0). Although the limits along the axes both exist and are equal to the value of the function at (0,0), [itex]\lim_{(z,t)\rightarrow (0,0)} f(z,t)[/itex] does not exist.

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So I still don't have a clue how to prove that x+y is continuous. Any more help?

what are you reading? Isn't this explained in every book on several variable calculus? i like courant, volume 2, and wendell fleming's book. basically you define the length of a vector |(a,b,c)| = sqrt(a^2+b^2+c^2), and do all the epsilon delta in those terms, word for word the same as in one variable.

It's differentiability where you have to do something new, because you can't divide vectors.

|x+y| ≤ |x| + |y|, implies continuity of addition.

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