Delta Epsilon in Three Dimensions?

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Could someone please briefly explain the thinking behind the Delta-Epsilon definition of a limit in three dimensions, Stewart doesn't seem to cover it?

answered May 16, 2017 by (4,220 points)
selected May 22, 2017 by Rowan

Are you interested in limits of a function, e.g. a curve in 3-dimensional space $$\mathbf{f}(t) = (x(t), y(t), z(t))$$ or a surface in 3-dimensional space $$\mathbf{g}(s,t) = (x(s,t), y(s,t), z(s,t))$$, or are you interested in the limit of a sequence, e.g. $$(x_n, y_n, z_n)_{n \in \mathbb{N}}$$?
I'll answer the first one. Recall that if $$f$$ is a function of one real variable, and $$a$$ is a cluster point (or limit point) of its domain, then we say $$\lim_{x \to a} f(x) = L$$ if for all $$\varepsilon > 0$$ there exists some $$\delta > 0$$ such that $$|x - a| < \delta$$ implies $$|f(x)-L| < \varepsilon$$.
Now suppose $$\mathbf{f}(t) = (x(t), y(t), z(t))$$ is a curve in space, and $$a$$ is a cluster point of its domain. It makes sense that if $$\mathbf{f}$$ takes values in $$\mathbb{R}^3$$ then its limits should be in $$\mathbb{R}^3$$. The easiest way to define a limit, in this case, is to say that $$\lim_{t \to a} \mathbf{f}(t) = (\ell_1, \ell_2, \ell_3)$$ if all three of the following are true:
$$\lim_{t \to a} x(t) = \ell_1$$
$$\lim_{t \to a} y(t) = \ell_2$$
$$\lim_{t \to a} z(t) = \ell_3$$
Alternatively, the same can be achieved by using the same $$\varepsilon-\delta$$ definition, but replacing the $$|\cdot|$$ with an appropriate notion of distance in $$\mathbb{R}^3$$ (known as a norm). The usual distance formula will do:
$$\Vert(x_1, y_1, z_1)-(x_2,y_2,z_2)\Vert = \sqrt{(x_1-x_2)^2+(y_1 - y_2)^2 + (z_1-z_2)^2}$$
Many other norms (in fact all norms on $$\mathbb{R}^n$$) are equivalent for this purpose.