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Delta Epsilon in Three Dimensions?

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asked May 14 in Calculus and Pure Mathematics by Daniel (160 points)

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?



1 Answer

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answered May 16 by simon_rigby (4,220 points)
selected May 22 by Rowan
 
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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:
\(\lim_{t \to a} x(t) = \ell_1\)
\(\lim_{t \to a} y(t) = \ell_2\)
\(\lim_{t \to a} z(t) = \ell_3\)
Or in other words, the limit is just defined pointwise (it is the limit in each of the three components simultaneously). We get all the usual facts in this way (for example, if a limit exists at a point, then it is unique).

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 onĀ \(\mathbb{R}^n\)) are equivalent for this purpose.
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