T.1) Formal definition of f[x]

We observed that the limit as x goes to infinity of f[x] gives horizontal asymptotes of the graph of f[x]: i.e. it identifies a constant L such that if x is large enough, f[x] will be as close as we like to L. We can quantify all of this:

Saying that we can get as close as we wish to L with f[x] says that given any (the Greek letter epsilon is often used in mathematics to indicate a small number used as a tolerance) we can make
| f[x] - L | <
This not only tells us that we have gotten close to L with f[x], it tells us how close we have gotten. Saying that x is sufficiently large can be quantified by saying there is a number N such that x > N.
We can then state the definition of the limit of f[x] as x goes to infinity:

Definition: The limit of f[x] as x goes to infinity is L if and only if
for any > 0, there is an N such that
if x > N then | f[x] - L | < .

T.1.a)

Take f[x] = (3 x^2 - 1) / (x^2 + x + 1). A quick plot shows that
y = 3 is a horizontal asymptote:

Clear[f];
f[x_]=(3 x^2 - 1)/(x^2 + x + 1);
Plot[f[x],{x, -1000, 
1000},PlotRange->All,PlotStyle->{{Red,Thickness[0.01]}}]

T.1.a.i)

Because we see y = 3 is the asymptote as x ->+, we can write
f[x] = 3. Given = 0.001, what is the appropriate N in the definition of f[x]? In other words, how big must x be in order to insure that | f[x] - 3 | < = 0.001?
Confirm the answer with a plot.

Answer:

Since we are interested in the f[x], we can concentrate on large positive values for x. Let's re-examine the plot for f[x] over the interval, [0, ):

Plot[f[x],{x, 0, 100},AxesOrigin->{0,0}]

By examining this graph, we see that for large x, f[x] is increasing and does not exceed 3.
To find the N corresponding to a given we just solve for where f[x] = 3 - . Here's the appropriate solve command for = 0.001:

epsilon=.001
NSolve[f[x]==3.- epsilon,x]

Now we use the largest of the answers given for the desired N. In other words, if N = 3000.33, then | f[x] - 3 | < 0.001.

We can check that this choice of N gives us values close enough to 3 with a graph.

Plot[{f[x],3+epsilon,3-epsilon},
	{x,3000.33,5000},
    
PlotStyle->{{},{Red},{Red}}]

The lines y = 3 + and y = 3 - can be thought of as an -tube. We know we have a correct choice for N since the graph of f[x] lies entirely inside the -tube for x > N.

T.1.a.ii)

Coninue with f[x] = (3 x^2 - 1) / (x^2 + x + 1).
Given = 0.00005, what is the appropriate N in the definition of
f[x]? In other words, how big must x be in order to insure
that | f[x] - 3 | < = 0.00005?
Confirm the answer with a plot.

Answer:

All we have done is changed the value of . We can use the same reasoning as in the solution to T.1.a.i). To find the N corresponding to = 0.00005, we solve for where f[x] = 3 - . Here's the appropriate solve command for = 0.00005:

epsilon=.00005;
NSolve[f[x]==3-epsilon,x]

Now we use the largest of the answers given for the desired N. In other words, if N = 60000.3, then | f[x] - 3 | < 0.00005.

We can check that this choice of N gives us values close enough to 3 with a graph.

Plot[{f[x],3+epsilon,3-epsilon},
	{x,60000.3, 150000},
    
PlotStyle->{{},{Red},{Red}}]

We have a correct choice for N since the graph of f[x] lies entirely inside the -tube for x > N.
Note that in general, changing the value of also changes the value of N. As a general rule, the smaller is, the larger N must be.

T.1.b)

Take f[x] = e^(-x/100) Sin[x]. A quick plot shows that y = 0 is a horizontal asymptote as x->:

Clear[f];
f[x_]=E^(-x/100) Sin[x];
Plot[f[x],{x, 0, 
1000},PlotRange->All]