Resolving Limits Graphically Near Domain Of Discontinuity Via One-Sided Limits

A "fun" math article by Jack Giffin on 9/30/2019! Yay math!

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The Problem

You have a function f(x) and a function g(x) such as is shown below.


We readily know that x→0f(x)=-4, but how do we find x→0g(f(x))? With direct substitution, we get x→0g(f(x))=x→-4g(x)=DNE (DNE means Does Not Exist) because x→-4g(x)=-1 and x→-4g(x)=1, so x→-4g(x)≠x→-4g(x). Let us analyse what happens in function composition.

The Solution Visually

Let us analyze the composition of g(f(x)). Comprehend that y=f(x) such that the domain—all x-values that the function passes through—is the input and the range—all y-values that the function passes through—is the output of the function. Now, imagine g(f(x))=g(y). From this perspective, the range outputted by f(x) becomes the new domain of g(f(x)). Observe the visual below. (Note that the grid is rotated 90 degrees to aid comprehension, so the x/y axes are flipped such that y is horizontal and x is vertical).

y=f(x); g(f(x))=g(y)

As the above image reveals, the range of f(x) is never less than -4, so the numbers going into g(y) are never less than -4. Thus, we can ignore all values of g(y) that are less than -4, leaving g(x) with only a single y=1 horizontal line which is the only remaining possible answer to x→0g(f(x)), so x→0g(f(x))=1 wins by default. Below is a graph of g(f(x)). Notice how its range consists of a single number: 1. In other words, g(f(x))=1 is true for each and every real value of x except for 0 which is undefined.


The Solution Mathematically

Now, how to do x→0g(f(x)) mathematically? Here's how:

  1. Let x→cK(L(x)) represent the situation where K(x) and L(x) are both arbitrary functions that both pass the vertical line test.
  2. Partial substitution with K(x→cL(x)) must have failed, else you would use this partial-substitution value as the result.
  3. There must be a local minima or a local maxima at L(c) such that either it's a local maxima due to L(c)-1/ ≤ L(c) ≤ L(c)+1/ or it's a local minima due to L(c)-1/ ≥ L(c) ≤ L(c)+1/. In other words, the function must have either both higher or both lower values on either side of c. If there is no local minima/maxima, then x→cK(L(x))=DNE. Recall that infinitely small values can be used to prove one-sided limits such as x→21+x/x-2=x→21+x/x→2x-2=2+1/x→2(x-1/)-2=3/2-1/-2=3/-1/=-3∞=-∞. These infinitely small values are also used with local minimas and local maximas in the same way. By an alternative definition, take the derivatives—rate of change at every single point—of L(x) and there exists either a local minima or a local maxima at c if L'(c)=0.
  4. If there is a local minima—increasing values on either side—at c, then x→cK(L(x))=y→L(c)K(y). Else, if there is a local maxima—decreasing values on either side—at c, then x→cK(L(x))=y→L(c)K(y).