B.1) Limits as x goes to infinity and horizontal asymptotes

We've already discovered that the dominant terms in a function, f[x], control the basic global shape of the graph.

If f[x] = L, the global shape on the right hand side of the graph
is the line, y = L.

If f[x] = M, the global shape on the left hand side of the graph is the line, y = M.
We call the lines, y = L and y = M horizontal asymptotes of f[x].
If f[x] = f[x] = L, then y = L is a two sided asymptote. This situation occurs whenever the dominant terms in f[x] are the same for x -> and x -> -

B.1.a)

B.1.b)

B.1.c.i)

B.1.c.ii)

These three problems show that we can find asymptotes by looking at graphs in very large scale. Unfortunately, this does not suffice, since we never know what scale is sufficiently large. We need a more precise definition. The precise definition is developed in the Tutorials notebook.

B.1.d)

Note that Mathematica knows how to find limits of the form,
for many types of functions:

Here's Mathematica 's computation for
(3 x^2 -1)/(x^2 + x + 1)

Clear[x];
Limit[(3 x^2 -1)/(x^2 + x +1), x->Infinity]

If we give a name to the function, we can also use:

Clear[f];
f[x_] = (3 x^2 -1)/(x^2 + x +1); 

Limit[f[x],x->Infinity]

Here's Mathematica 's computation for
Sin[x]/e^(x/100)

Limit[Sin[x]/E^(x/100),x->Infinity]

In this case Mathematica has guessed what the limit is by trying very large values of x--essentially this is the same as looking carefully at a big piece of a graph.

Here's Mathematica 's computation for
(3 x -1)/Sqrt[x^2 + x + 1]

Limit[(3 x -1)/Sqrt[x^2 + x +1],x->Infinity]

Here's Mathematica 's computation for
(3 x -1)/Sqrt[x^2 + x + 1].

Limit[Clear[f];
f[x_]=(3 x -1)/Sqrt[x^2 + 
x+1],x->-Infinity]