3.2.2
3.2.2
FINDING LIMITS FROM A GRAPH
Consider a function f(x) on the graph at right,
with a point on the x-axis designated as 'c'.
y
f(x)
x
c
The point on the graph corresponding to
the x-value 'c' is "f(c)". Thus f(c) represents
a height, or a y-value, on the graph. We are
going to give this y-value another name, "L".
y
f(x)
L = f(c)
x
c
Notation
In order to discuss limits, we introduce a new kind of notation.
lim f(x) = L
x→c
is read: As x approaches c, the limit of f(x) is L.
When a small " – " is added as a superscript to "c" we have:
lim f(x) = L which is read: As x approaches c from the left the limit of f(x) is L.
x→c Note that the small "-" superscript has no affect on whether "c" is positive or negative.
When a small "+" is added as a superscript to "c" we have:
lim f(x) = L which is read: As x approaches c from the right the limit of f(x) is L.
x→c+
Again, the small "+" superscript has no affect on whether "c" is positive or negative.
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1
3.2.2
Reading Limits from a Graph
If you have the graph of a function, you can find a limit using the three-step process below.
1) Find the x-values.
2) Find the corresponding y-values on the graph.
3) Find the height that the y-values are getting closer and closer to,
or for those who prefer Latin grammatical structures to Germanic:
3) Find the height to which the y-values are getting closer and closer.
Consider the graph below.
y
8
1
2
3
4
f(x)
Step 1: To find
lim f(x) we first note that the 'x's are approaching 2 from the left.
x→2 y
8
1
2
3
4
f(x)
Notice that the 'x's are zooming in toward the '2' from the left side.
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3.2.2
Step 2: We now trace each 'x' upward or downward until we find it's corresponding y-value
on the graph.
y
8
1
2
3
4
f(x)
Step 3. As the 'x's get closer and closer to '2' from the left side, the y-values are climbing up
closer and closer to a height, or y-value, of 8. Imagine a tiny hiker climbing up the graph on the
corresponding y-values. As the hiker's horizontal position gets closer and closer to 2, he expects
to get closer and closer to a height of 8. In other words, as the value of x approaches '2' from the
left, the hiker's expected height, or y-value approaches '8'.
y
8
1
2
3
4
f(x)
Now consider the limit as 'x' approaches '2' from the right: lim f(x)
x→2+
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3
3.2.2
Step 1. The x's get closer and closer to '2' from the right, i.e. the 'x's are zooming in on '2' from
the right.
y
8
1
2
3
4
2
3
4
f(x)
Step 2: Find the corresponding y-values.
y
8
1
f(x)
Step 3: Once again, as the hiker's horizontal position approaches '2' from the right, he expects to
reach a height of '8', or we could say: as the value of 'x' approaches '2' from the right, the value
of y approaches '8'.
Since the lim f(x) (which was 8) equals the lim f(x) (which was 8 again), then we have
x→2 x→2+
lim f(x) = 8 .
x→2
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3.2.2
In words, since the limit of f(x), as 'x' approaches '2' from the left equals 8, and the limit of f(x)
as 'x' approaches '2' from the right equals 8, then the limit of f(x) as x approaches 2 equals 8.
In general lim f(x) can only exist when lim f(x) = lim f(x).
x→c
x→cx→c+
If lim f(x) = L and lim f(x) = L, then lim f(x) = L.
x→c x→c+
x→c
The limits will exist when the graph is continuous. A graph is continuous at x0 if the expected
height equals the actual height, that is if:
lim 𝑓(𝑥) = f(x0)
𝑥→𝑥0
Now consider the function below, which is similar to the previous function. This one, however,
has a missing point at x = 2. It is not continuous at xo.
y
8
1
2
3
4
f(x)
Once again find the limit as 'x' approaches '2' from the left, i.e. lim f(x).
x→2Remember to use all three steps.
Step 1: Find the 'x's.
y
8
1
2
3
4
f(x)
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5
3.2.2
Step 2: Find the corresponding y-values.
y
8
1
2
3
4
f(x)
Step 3: Find the height to which the hiker, sic y-values, are getting closer and closer.
y
8
1
2
3
4
f(x)
Note that even though there is a missing y-value corresponding to x = 2, the hiker is still getting
closer and closer to a height of 8, so we have a limit as x approaches 2 from the left,
i.e. lim f(x) = 8.
x→2If one were to ask, maybe the hiker, What exactly is the height at x = 2?, or in mathematical
terms, What is f(2)?, we would have to say f(2) does not exist (DNE) because there is no
corresponding y-value for an x-value of 2. So even when f(x0) doesn't exist, there can still be a
limit.
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3.2.2
Similarly, using the three steps we could say lim f(x) = 8.
x→2+
y
8
1
2
3
4
f(x)
Finally, we can say that since lim f(x) = lim f(x) = 8, we can say lim f(x) = 8.
x→2x→2+
x→2
Now consider another similar graph. This one also has a discontinuity at x = 2, but note that
on the graph below, f(2) = 10. But what about lim f(x)?
x→2
Again, we'll have to use the 3 steps and we must check both the limit as x approaches 2 from the
left and the limit as x approaches 2 from the right, i.e. lim f(x) and lim f(x).
x→2x→2+
If these are equal, then we have the lim f(x).
x→2
y
10
8
1
2
3
4
f(x)
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3.2.2
Step 1: For lim f(x), we find the x-values.
x→2Step 2: We find the corresponding y-value, or heights, on the graph.
Step3: We find the height that the hiker, or y-values are getting closer and closer to.
y
10
8
1
2
3
4
f(x)
Note that our hiker is getting closer and closer to a height of '8'. He can't fly. He must stay on
the path arranged by the corresponding y'values as x approaches 2.
Similarly, as c approaches 2 from the right, the hiker gets closer and closer to a height of 8, or
the corresponding y-values approach a height of 8. So even though f(x) = 10, we have:
lim f(x) = lim f(x) = 8. Therefore lim f(x) = 8.
x→2x→2+
x→2
Using the graph above, let's find:
lim f(x). We'll start with lim f(x).
x→0
x→0-
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3.2.2
Step 1: Find the 'x's.
y
10
8
1
2
3
4
f(x)
Step 2: Find the corresponding y-values, or heights.
y
10
8
1
2
3
4
f(x)
Step 3: Since we are approaching '0' from the left, the corresponding y-values are getting closer
and closer to, or actually staying at, a height of -6. The hiker expects to be at a height of 8
when he reaches the horizontal distance x = 0.
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3.2.2
Now we consider lim f(x).
x→0+
Step 1: Find the 'x's.
y
10
8
1
2
3
4
f(x)
Step 2: Find the corresponding y-values, or heights.
y
10
8
1
2
3
4
f(x)
Step 3: This time a tiny hiker would approach a height of -6. Thus lim f(x) = -6.
x→0+
Since lim f(x) = lim f(x) = -6, we can conclude that a) lim f(x) = -6, and
x→0x→0+
x→0
b) there are no clip art hikers going downhill. We can also see that f(0) = -6.
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3.2.2
For the same graph, what about lim f(x)?
x→∞
Step 1: Infinity is the farthest distance to the right on the x-axis. We can't draw a line going
to the right forever in order to find the x-values as 'x' approaches infinity, but fortunately,
we come equipped with minds that can imagine such a process. We'll just pinpoint some
'x's on their way toward ∞.
y
10
8
∞
-∞
1
2
3
4
f(x)
Step 2. Next, we imagine where the corresponding y-values would be.
y
10
8
∞
-∞
1
2
3
4
f(x)
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3.2.2
Step 3: We can see that as 'x' approaches infinity, the corresponding y-values are going down
and down forever. They never level off or get closer and closer to one particular value
height, or y-value. We say: lim f(x) = -∞.
x→ -∞
While "-∞" is not an actual limit, more like the lack of a limit, it does tell us some important information about the function f(x); that is, that f(x) drops forever as x approaches
positive infinity.
What about lim f(x) ?
x→ -∞
Step 1: Find the x-values. Again, we can only imagine them since we can't draw a line going
forever to the left. We'll just pinpoint a few of the 'x's on their way to forever left.
Step 2: We can also imagine the corresponding y-values. If f(x) continues to be a horizontal
line as 'x' approaches -∞, then the corresponding y-values will all lie on the line at a
height of -6.
y
10
8
∞
-∞
-
1
2
3
4
f(x)
Step 3: The corresponding y-values are getting closer and closer to, or actually, staying at
a height of -6. The hiker expects to stay at a height of -6. Hence lim f(x) = -6.
x→ -∞
Remember: Limits are always y-values, i.e. heights on a graph.
Exercise: Find lim f(x)
x→4
Answer: Since lim f(x) = lim f(x) = -1, lim f(x) = -1.
x→4x→4+
x→4
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3.2.2
Now consider the graph of the function: f(x) = - 3│x-2│
x-2
y
f(x)
x
Consider lim f(x). Using the three steps, we see that lim f(x) = 3.
x→2x→2Now consider lim f(x) (below). Using the three steps, we see that lim f(x) = -3.
x→2+
x→2+
y
f(x)
x
Since lim f(x) = 3 ≠ lim f(x) = -3,
x→2x→2+
the
lim f(x) does not exist.
x→2
Whenever there is a break in the graph at x = 'c', such as that at x = 2 above, the limit as 'x'
approaches 'c' will not exist because the limit as 'x' approaches 'c' from the left will not equal the
limit as 'x' approaches 'c' from the right.
Note that though the limit could exist when the discontinuity was a hole, the limit does not exist
when the discontinuity is a break. At breaks, the limit as x approaches c from the right doesn't
match the limit as x approaches c from the left.
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3.2.2
Finally, consider the graph of the function f(x) = 1/x.
asymptotes.
The x-axis and the y-axis are both
y
x
If we use the three steps, we can determine that lim
1 = 0, because as the 'x's approach
x→∞ x
infinity going forever to the right, the hiker expects to reach a height of 0, and the corresponding
y-values get closer and closer to a height, or y-value of '0'. (See right side of diagram.)
y
x
What about lim 1 ?
x→ -∞ x
Using the three steps, we can determine that lim 1 also equals
x→ -∞ x
zero, because as the 'x's approach negative infinity going left forever, the corresponding y-values
also get closer and closer to a height, or y-value, of '0'. (See left side of diagram.) Or in
mathematical terms:
lim 1 = 0.
x→ -∞ x
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