Epsilon-Delta Limits

Limits required a rigorous definition, and the epsilon-delta definition satisfied this need.

This definition states that for `x` in a subset of the real numbers arbitrarily near a value `c`, `\displaystyle\lim_{x\to c}f(x)=L` if and only if
`\forall \epsilon>0, \exists \delta: 0<\lvert x-c \rvert<\delta\implies\lvert f(x)-L\rvert<\epsilon`.

Note that `f(x)` need not be defined at `c`!
Proofs that rely upon the use of the minimum function artificially place a maximum on the distance from `c`
and restrict the domain, but this is still valid because the domain can be arbitrarily small by definition.

Example 1:

Prove `\lim_{x\to\pi}(x)=\pi`.
Start by finding `\epsilon`, then define `\delta` in relative terms.
`\epsilon > \lvert f(x)-\pi \rvert = \lvert x-\pi \rvert>0`

Choosing `\delta=\epsilon` completes the proof.

Example 2:

Prove `\lim_{x\to1}(5x-3)=2`.
Once again, start by finding `\epsilon`, then define `\delta`:
`\epsilon > \lvert f(x)-2 \rvert = \lvert (5x-3)-2 \rvert = \lvert 5(x-1) \rvert = 5\lvert x-1 \rvert>0\implies\frac\epsilon5 >\lvert x-1 \rvert>0`

Choosing `\delta=\frac\epsilon5` completes the proof.

Example 3:

Prove `\lim_{x\to7}(x^2+1)=50`.
Using `\delta > \lvert x-7 \rvert > 0`:
`\lvert f(x)-50 \rvert = \lvert (x^2+1)-50 \rvert = \lvert x^2-49 \rvert = \lvert x-7 \rvert\lvert x+7 \rvert < \delta\lvert x+7 \rvert`
Assume `\delta \leq 1 \implies \lvert x-7 \rvert\leq1 \implies -1 \leq x-7 \leq 1 \implies 6 \leq x\leq 8 \implies \lvert x\rvert \leq 8`.
TODO: Link triangle inequality
By the triangle inequality, `\lvert x+7 \rvert \leq \lvert x \rvert + 7 \leq 15`. Hence, `\lvert f(x)-50 \rvert < 15\delta`.

Choosing `\delta=\min(1,\frac\epsilon{15})` completes the proof and satisfies the assumption.

TODO: Infinite limits
TODO: Multivariate limits, or a link to a multivariate calculus counterpart

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Information and examples taken from Brilliant.