B.3) Squeezing limits

B.3.a.i)

We can use a squeeze to find limits as x->a under special conditions.
For example, take f[x] = x Sin[1/x]. Explain why f[x] = 0.

Answer:

Clear[f];
f[x_]= x Sin[1/x]

Here we always have f[x] between x and -x (or, so we don't have to worry about which one is on top, -|x| <= f[x] <= |x| ). Now we make several plots of the three functions, -|x|, f[x], and |x|. (|x| is written as Abs[x] in Mathematica and is shown in blue, while -|x| is written as -Abs[x] and is shown in red.)

Plot[{-Abs[x], f[x], Abs[x]}, {x,-10,10},
	
PlotRange->{-10,10},
	PlotStyle->{{Red},{},{Blue}}]

Plot[{-Abs[x], f[x], Abs[x]},{x,-1,1},
	
PlotRange->{-1,1},
	PlotStyle->{{Red},{},{Blue}}]

Plot[{-Abs[x], f[x], Abs[x]},{x,-.1,.1},
	
PlotRange->{-.1,.1},
	PlotStyle->{{Red},{},{Blue}}]

Plot[{-Abs[x], f[x],Abs[x]},{x,-.001,.001},
	
PlotRange->{-.001,.001},
	PlotStyle->{{Red},{},{Blue}}]

Looking at these three functions in different scales we notice that the picture is complicated near 0, but it is clear that f[x] is always trapped between the graphs of |x| and -|x|. Since both of these graphs go through the point {0, 0}, the graph of f[x] is "squeezed" into approaching the point {0, 0}, even though f[0] is undefined. Thus, it is clear that f[x] is 0.

B.3.a.ii)

Continue with f[x] = x Sin[1/x]. Does the point {0, 0} actually lie on the graph of f[x]? Explain your answer.

Answer:

The point {0, 0} does not lie on the graph of y = f[x] because f[0] is not defined. When we ask Mathematica to evaluate f[0], this is what we get:

f[0]

This says that Mathematica has discovered that when we plug 0 into the formula for f[x], we get 0/0, which is an indeterminate---i.e., f[0] is not defined because it results in a division by 0. Mathematicians describe this situation, f[x] = L but f[a] does not, by saying that the graph of f[x] has a "hole" at the point {a, L}