Again let us start by looking at some examples. These should illustrate how limits help us fill in missing values (like the ones we want when finding instantaneous rate of growth).
Take f[x] = (x^2 - 1)/(x -1).
Clear[f];
f[x_]=(x^2 -1)/(x-1)
Note that f[1] makes no sense, since f[1] = 0/0, which is undefined. Mathematica knows this fact too:
f[1]
Now look at a truescale plot of f[x] over 0 <= x <= 2:
Plot[f[x],{x,0,2},PlotRange->All,
AspectRatio->Automatic]
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr57.gif]](LimitsBasicsgr57.gif)
Notice that in this case the graph looks just like that of the line y=x+1. (Except that on some computers, you may get a warning that we were dividing by 0 at x=1.) Explain this fact.
Continue with f[x] = (x^2 -1)/(x -1).
Explain why
f[x] = 2.
Three points are worth mentioning here:
First, in evaluating
f[x], it does not matter what f[a] is, or even if f[a] makes any sense, it only matters what the values of f[x] are for x's that are very near a.
Second, reasonably nice functions (like polynomials) should have
f[x] = f[a]. Such functions will be called continuous at a. In our example f[x] is not continuous at x = 1 (since f[1] does not exist), but g[x] = x+1 is continuous at x = 1.
Third, when an algebraic simplification can be used to eliminate division by 0 all that is being changed in the definition of the function is the definition at the point where division by zero occurred, not the values for nearby points where both the original function and the simplified function are both defined. These sorts of operations should not change the limit.
This example shows that even if such algebraic simplification is not available we may get a limit. It also shows how some functions look progressively simpler under higher and higher magnification:
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr70.gif]](LimitsBasicsgr70.gif)
Take f[x] = Sin[x]/x. Make several plots over smaller and smaller x-intervals that contain x = 0. What do the plots say about
f[x]?
Clear[f];
f[x_]=Sin[x]/x
Look at the following plots, which are done over increasingly smaller ranges of x-values.
Plot[f[x],{x,-10,10},PlotRange->{-.5,1.5}]
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr76.gif]](LimitsBasicsgr76.gif)
Plot[f[x],{x,-1,1},PlotRange->{0,1.5}]
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr78.gif]](LimitsBasicsgr78.gif)
Plot[f[x],{x,-.1,.1},PlotRange->{0,1.5}]
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr80.gif]](LimitsBasicsgr80.gif)
Plot[f[x],{x,-.001,.001},PlotRange->{0,1.5}]
![[Graphics:LimitsBasicsgr2.gif]](LimitsBasicsgr2.gif)
![[Graphics:LimitsBasicsgr82.gif]](LimitsBasicsgr82.gif)
Looking at higher and higher magnification allows us to look at values of x that are closer and closer to 0. The graphs all appear to go through the point {0, 1}, and the last graph shows that if
-0.001 <= x <= 0.001, we can't tell the difference between f[x] and 1. It appears that
f[x] is 1. This is also one that Mathematica knows how to do:
Limit[f[x],x->0]