Limits
Whoever invented the term “limit” clearly was no expert in English. It really should be
called “guess,” which is really what a limit is. Consider the following limit, for example:
What this really says is: “Guess what 1/x gets close to as x gets .
really large (approaches infinity).”
Similarly, the following limit asks “What does the value of (x2 – 1) / (x + 1) get close to
as x approaches infinity?”
Note: The arrow in this limit indicates what value x
approaches. So, in both these cases, x gets close to infinity.
To solve a limit, you need to take a guess at this value. Of course, there are methods to
investigate the situation between making an arbitrary guess. The method you use depends
on the problem at hand. The following is a method that can be used to find any limit
(though it is terribly inefficient). It’s a good way to get a feel for limits.
Method 1: Find a pattern
x=
1/x=
1
1
2
0.5
5
0.2
10
0.1
20
0.05
50
0.02
100
0.01
500
0.002
1000 10000
0.001 0.0001
In this table, I choose values of x and then plug those values into the function F(x) = 1/x.
As x gets really large, the value of 1/x gets closer to 0. Therefore, we say that the limit of
1/x as x goes to infinity is 0. Check out the following graph:
This is a graph of F(x) = 1/x
Notice how the line gets really close to the x-axis. As
the value of x gets larger, the value of the F(x) gets
really small. If you put a monstrous number into F(x),
it will return a number that is approximately equal to
0. Thus, the limit, or the “guess,” of 1/x as x goes to
infinity is 0.
So, what if the limit does not equal 0? What if the line on the graph doesn’t get closer to
the x-axis? The process for solving the limit is exactly the same. Let’s take a look at
another limit presented earlier in this text:
To solve this limit, we can again choose larger and larger values
for x and see if we can find a pattern (i.e. guess what the function
becomes as x approaches infinity)
(x2-1)
x=
/ (x+1) =
1
0
2
1
5
4
10
9
20
19
50
49
100
99
500
499
1000
999
10000
9999
Hmm, doesn’t seem like it’s going to 0. In fact, the value seems to always be one less
than x. We’ll discuss why this is later. Right now, we need to take our guess. If the value
of the function is always one less than x and x is approaching infinity, we can
legitimately guess that the function also approaches infinity. Therefore…
Let’s look at one more…
Ok, so let’s choose some x’s…
x=
(2x-1) / (x+1) =
1
0.5
5
1.5
10
1.727
50
1.941
100
1.970
500
1.994
1000
1.997
10000
1.9997
Of course, it doesn’t matter what values for x you choose. These are completely arbitrary.
I am simply trying to learn what happens to the function as you put in larger and larger
values of x. It seems that the function gets really close to equaling 2. Thus…
We’ve looked at three examples of limits, each giving a different result. It’s clear that
limits can equal any value. Here’s a challenge: Find the following limit…
So far, we’ve covered limits that ask us to guess what happens to an function as x
approaches infinity. It should be mentioned that finding a limit in which x approaches
negative infinity requires very much the same steps for positive infinity. The only
difference is that instead of choosing larger positive numbers for x, you would choose
values for x that approach negative infinity (i.e. larger and larger negative numbers).
But, what if the limit asks you to guess what happens when x approaches something other
than infinity? In other words, what happens if the number after the arrow is a number
rather than the infinity symbol? In this case, three things can happen.
Case 1 – Simple Substitution
This limit asks us to guess the value of 1 / (x – 3) as x gets close
to 4. In this case, a guess is easy. Just plug 4 in for x and get
your answer. So, 1 / (4 – 3) = 1 and so the limit is equal to 1.
Case 2 – Division by Zero
This limit asks us to guess the value of 1 / (x – 3) as x
approaches 3, which is much trickier. We can’t simply plug in 3
this time since the result would be 1 / (3 – 3) = 1 / 0, which is
invalid since nothing can be divided by zero.
To solve this limit, we have to choose values for x that get closer and closer to 3, but that
never exactly equal x. So, let’s bring back that table…
x=
1 / (x – 3) =
1
-0.5
2
-1
2.6
-2.5
2.8
-5
2.9
-10
2.99
-100
2.999
-1000
2.9999
-10000
As x gets closer to 3, the function seems to go to negative infinity. So, you might think
that the limit as x approaches 3 is equal to negative infinity. But this limit is deceptive…
The values we chose for x approach 3 from the negative direction. In other words, all our
values were less than 3 and approached 3 from the left side of the number line. What
happens if we choose x values that get closer to 3 but are all larger than 3? Values that
approach x from the right side of the number line?
x=
1 / (x – 3) =
5
0.5
4
1
3.4
2.5
3.2
5
3.1
10
3.01
100
3.001
1000
3.0001
10000
Approaching 3 from the right side shows that the function goes to positive infinity rather
than negative infinity. So which is it? Positive or negative? Let’s take a look at the graph:
This is a graph of F(x) = 1 / (x – 3)
Ignore the vertical line in the center…it’s
just connecting two points, but it doesn’t
actually exist. In reality, the two prongs of
this function continue up and down
infinitely and never connect. As x
approaches 3 from the negative side (red
line), the graph of F(x) drops into negative
infinity. As x approaches 3 from the
positive side (blue line), the graph rises to
positive infinity.
When we say that the function goes to positive infinity as x approaches 3 from the
positive side, we can write that as the following:
Notice the plus sign after the 3. This indicates that x
approaches 3 from the positive (right) side.
Conversely, when we say that the function goes to negative infinity as x approaches 3
from the negative side, we can write that as the following:
The negative symbol after the 3 indicates that x
approaches 3 from the negative (left) side.
As we can clearly see, approaching 3 from two different directions gives us two
completely different values. Now the tough question…what’s the limit of this function as
x approaches 3? Remember that a limit is simply a question asking you to guess the value
of an function when x equals 3. Approaching 3 from the left side gets negative infinity
and approaching from the right side gets positive infinity. Good luck taking a guess!
In this case, we say that the limit does not exist (i.e. a guess cannot be taken). This also
means that in order for a limit to exist, approaching the limit value from the negative and
positive directions must yield the same answer.
The reason this limit does not exist is because when x approaches 3 from two different
directions, the limit produces two different results. So, is there an example of a limit that
exists and has zero in the denominator? There most certainly is:
Let’s look at the graph…
Approaching zero from the positive
direction (blue) leads us to believe that the
limit is equal to positive infinity.
Approaching from the negative (red)
direction also indicates a limit that equals
positive infinity.
Two different directions yielding the same
result? This sure makes our guess a lot
easier to make! If x equals 0, then 1/x2 must
equal positive infinity.
If approaching the limit value (which in this case is zero) from two directions yields the
same answer, then the limit exists (i.e. we can logically guess the value of the function
when x equals zero). In this case, the limit equals positive infinity:
When zero appears in the denominator of a fully simplified function, you will always
have a vertical asymptote. An asymptote is an imaginary line that the function can never
cross. The asymptote for the above limit is the line x = 0 (the y-axis). As you will notice,
the red and blue lines get really close to the y-axis, but never actually cross it. Again,
asymptote’s appear whenever there is a division by zero in a fully simplified function.
The following are two examples of limits with a division by zero:
The vertical asymptote for the above limit is x = 1. In other words, x can never equal 1 in
the graph because when x = 1, y = 1 / 0 and dividing by zero is not possible.
The vertical asymptote for the above limit is x = 2. So, x can never equal 2. The graph
will get really close to x = 2, but the graph will never cross x = 2.
If you have trouble understanding why the graphs do what they do (i.e. go to infinity), try
constructing tables like at the beginning of this sections. Choose values for x that get
really close to the limit value (1 for the first limit and 2 for the second limit). What
happens to the function as you replace x with values that get closer and closer to the limit
value?
As stated before, vertical asymptotes appear whenever a fully simplified function has a
division by zero at some point. But, what if the function is not fully simplified?
Case 3 – Simplifiable Function
Is that even a word? Simplifiable? It is as far as I’m
concerned! Limits in this category require you to alter the
function before finding the solution.
Method 2: Modify the function
Method 1 (finding a solution) would work just fine for this problem. But, method 1
(creating tables and finding patterns) is terribly inefficient. For this limit, there is a much
easier way of guessing what the function equals as x gets close to 1. Of course, the reason
we can’t calculate (x2-1) / (x-1) is because when x equals 1, the denominator is equal to
zero. So, we must find a way to get rid of this troublesome denominator.
The way to do this is to find something we can cancel from the top and bottom of this
function. If you are unfamiliar with fractions or you are unfamiliar with factoring
polynomials, you’ll want to read the section entitled “Fractions” and “Factoring,” which
can be found on the website www.mathtechniques.com/algebra.html.
Once you understand factoring and how to manipulate fractions, you can take the
following steps to solve the limit:
Step 1 : Factor the polynomials in the nominator and denominator
Step 2 : Separate the fraction into the product of two fractions
Step 3 : Simplify each function  (x + 1) / 1 = (x + 1) and (x – 1) / (x – 1) = 1
Note: I separated the denominator into 1 and (x – 1) so it’s easier to see where (x+1) / 1
comes from. This whole thing could also be written without the 1’s…
So, (x2 – 1) / (x – 1) is the same as (x + 1).
We can replace the original function with (x + 1). And since
they’re both the same, the limit of both are the same. And, as
you can see, this limit is a lot easier to calculate since there is
no zero in the denominator. Just plug in 1 for x…
Thus, the original limit is also equal to 2.
The idea behind this method is eliminating the zero in the denominator. There are
different ways to do this. Yet, it always comes down to canceling something from the top
and bottom of the fraction. One of the more advanced implementations of this method is
a technique called “Multiplying by the Conjugate.” The “conjugate” of a polynomial is
anything that can be multiplied by that polynomial to produce a difference of squares.
The terms “difference of squares” and “conjugate” are also discussed in the section
entitled “Factoring.”
Multiplying by the Conjugate
This is a good example of a limit requiring the use of a conjugate.
Multiplying top and bottom by the conjugate gives the following:
Step 1 : After multiplying top and bottom by the conjugate, FOIL the denominator.
Step 2 : Cancel the (x – 1) from top and bottom.
And with that, the denominator has been eliminated and we no longer have a division by
zero problem. Taking the limit now is as simple as setting x equal to 1.
Since the function in this limit is
the same as the original, the answer
to this limit is also the same as our
original limit.
This concludes finding limits with limit values that are not infinity or negative infinity.
We have one final method to explore before deeming ourselves experts on limits.
Method 3 – Comparing Rates of Change
This method is a short-cut to doing Method 1. It allows you to predict the result of
making a table of results without actually making the table. And it’s all based on logic!
Let’s work with this limit, which you may recognize
from the beginning of this section. We’ve already
guessed that it equals 2 based on the tables we
constructed. Let’s see if we were right…
I’m going to start by constructing a table that tells me what the top and bottom equal as x
gets really, really huge.
x
2x – 1
x+1
10
19
11
100
199
101
1,000
1,999
1,001
10,000
19,999
10,001
100,000
199,999
100,001
1,000,000
1,999,999
1,000,001
10,000,000
19,999,999
10,000,001
As x approaches infinity, it seems that 2x – 1 gets closer and closer to twice of x + 1.
Consider x = 10,000,000, for example. 19,999,999 is extremely close to 20,000,002,
which is twice the value of 10,000,001. So, we can assume that when x approaches
infinity, 2x – 1 becomes twice as large as x + 1. If we divide 2x – 1 by x + 1, that’s like
dividing 2 by 1, in which case our answer would be 2.
An easier way to go about this is to find the rate of change of the nominator and the
denominator. Every time x increases by 1, 2x – 1 increases by 2. Thus, it’s rate of change
is 2. Every time x increases by 1, x + 1 increases by 1. Thus, it’s rate of change is 1. This
means that the top grows exactly twice as fast as the bottom and if we were to divide the
top by the bottom with a huge value for x, we’d get something really close to 2.
Also notice what happens to the 1’s in the limit as x approaches infinity. They become
less and less significant. When x = 10,000,000…
2x – 1 = 19,999,999 and
x +1 = 10,000,001
and
2x = 20,000,000
x = 10,000,000
 not a lot of difference!
So, a quick way to determine the limit is to consider only the fastest growing part of the
top and bottom of the fraction. For the top, the fastest growing part was the 2x. For the
bottom, the fastest growing part was the x. Let’s take a look at a few more limits.
The fastest growing term on the top is 4x.
The fastest growing term on the bottom is -2x.
4x / -2x = -2, so as x approaches infinity, this fraction
approaches -2. The limit, therefore is -2.
The fastest growing term on the top is x2.
The fastest growing term on the bottom is -2x2.
x2 / -2x2 = -1/2 = -0.5, so as x approaches infinity, this
fraction approaches -1/2.
Top : 4x3
Bottom : -2x3
4x3/-2x3 = -2
This last one may seem tricky with so many terms. Just remember that as x gets really
large, the x3 grows much faster than anything else. All the other terms (5x2, x, 3x, 1)
become insignificant when compared to x3. For example, what happens when x = 10,000?
4x3 + 5x2 + x = 4,002,500,010,000 and
-2x3 + 3x + 1 = -1,999,999,969,999 and
4x3 = 4,000,000,000,000
-2x3 = -2,000,000,000,000
Essentially, what we’re doing is estimating. Remember that limits are simply guessing
what happens when x gets to a certain value and estimating is guessing.
Graph of the previous limit.
Notice that as x gets larger, the
graph gets closer to y = -2. The
graph will never cross y = -2,
but it will get really close. Do
you remember what this is
called? If you said asymptote,
then you’d be right.
This is a horizontal asymptote and is defined by the line y = -2, since that is the line
which the graph approaches and never crosses. There will be a horizontal asymptote
whenever the degree of the numerator is equal to the degree of the denominator. The term
degree refers to the highest power in a polynomial. So for the above limit, the degree of
the nominator is 3. The degree of the denominator is also three, which is why there’s a
horizontal asymptote.
Now you ask, what if the degrees in the nominator and denominator are different? The
answer is easier than you would think. If you have a decent understanding of limits, the
following one should be pretty easy…
In this case, the top grows MUCH faster than the bottom. So,
when x is huge, the function will be a huge number divided by a
relatively small number. A huge number divided by a small
number is still somewhat of a huge number.
Let’s consider x = 100 and x = 10,000:
4(100)3 / 98 is close to 40,000
and
4(10,000)3 / 9,998 is close to 400,000,000
The degree of the nominator is 3 and the degree of the denominator is 1, which means
that the numerator far exceeds the denominator. As x approaches infinity, this function
becomes incredibly large, meaning that the limit is equal to infinity. The only time that a
horizontal asymptote does not exist is if the degree of the nominator is larger than the
degree of the denominator.
Because this graph never crosses the
x-axis, you may think there’s a
horizontal asymptote at y = 0. But,
the definition of an asymptote is an
imaginary line that a graph gets
increasingly close to. As you can see,
this graph goes to infinity in both
directions and thus has no horizontal
asymptotes.
What if the degree of the denominator is larger than the degree of the nominator? In this
case, the denominator will be vastly larger than the nominator and, as we all know, a
small number divided by a huge number results in a fraction that is close to zero. The
following limit illustrates this point:
This is similar to the previous limit, except the
nominator and denominator have been switched. When
the bottom far exceeds the top, then the fraction gets
closer to zero as x approaches infinity.
And this concludes this section on limits.