Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics

1. 1.2 Finding Limits Graphically and Numerically

2. Objectives:
(1) Estimate a limit graphically or numerically.
(2) Identify ways a limit fails to exist.
(3) Find a limit using the formal definition.
An Introduction to Limits
Consider the function
f(x) =
x2
− 1
x − 1
.
The domain of the function is {x|x = 1}. The function is not defined for x = 1. What is
the nature of f when x is very close to x = 1? We can explore this question numerically by
using the table feature of our graphing calculator. The following figure is a screen capture
that shows the corresponding y-coordinates when x approaches x = 1 from the left, then x
approaches x = 1 from the right.
It appears that the closer x gets to x = 1, the closer the y-coordinate gets close to y = 2.
We can support this conjecture with a graph. The following is the graph of the function.
From the graph, it appears that when x is close to x = 1, the corresponding y-coordinate is
close to y = 2. What we can’t see in this screen capture is that there is a hole in the graph
at the point (1, 2).
This leads us to an informal notion of a limit. If f(x) becomes arbitrarily close (as close as
you can imagine) to a single finite number L as x approaches c from either side, the limit
of f(x), as x approaches c, is L. This is written
lim
x→c
f(x) = L.
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3. Example.
Use the table feature of your graphing calculator to estimate the following limit, Use the
graph to confirm your result.
lim
x→2
x − 2
x2 − 4
Solution.
Looking at the table, it appears that when x → 2, f(x) → 0.25.
The following is a small portion of the graph. The horizontal axis ranges form x = 0 to
x = 3 with a scale of 1. The vertical axis ranges from y = 0 to y = 1 with a scale of 0.25
As we can see, when x is close to x = 2, y is close to y = 0.25.
We will work several of problems 9-18 in class.
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4. Limits That Fail to Exist
Example (A jump in the graph).
Consider
lim
x→0
|x|
x
The following is the graph of the function near the origin.
We can see that f(x) = −1 for all x < 0 and f(x) = 1 for all x ≥ 1. When x → 1, f(x) does
not approach a single finite number. So, the limit does not exist.
Example (Unbounded Behavior).
Consider
lim
x→0
1
x2
.
We know there is a vertical asymptote at x = 0. The following is the graph of the function
near the origin.
We can see that when x → 0 from either the right or left, f(x) → ∞. So, the limit does not
exist.
Example (Oscillating Behavior).
Consider
lim
x→0
sin
1
x
.
As x → 0 from the right, 1
x
→ ∞. Likewise, as x → 0 from the left, 1
x
→ −∞. We should
expect our sine curve to have extremely rapid oscillations near the origin. Unfortunately,
The graphing calculator does not handle this graph very well. See figure 1.10 on page 51.
When x → 0, f(x) does not approach a single finite number. So, the limit does not exist.
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5. A Formal Definition of a Limit
Definition (Definition of Limit).
Let f be defined on and open interval containing c, except possibly at c, and L be a real
number. The statement
lim
x→c
f(x) = L
means for each > 0 (Geek lowercase epsilon), there exists δ > 0 (Greek lowercase delta)
such that if
0 < |x − c| < δ, then |f(x) − L| < .
This means that if we want the function to get arbitrarily close to L, closer than , we can
find a value δ such that if x is closer to c than δ, f(x) will be closer to L than . See figure
1.12 on page 52.
Example.
Use the definition of limit to prove that
lim
x→4
4 −
x
2
= 2.
Solution.
Let > 0 be given. We must find δ as a function of such that
4 −
x
2
− 2 < whenever 0 < |x − 4| < δ.
|f(x) − L| < ⇒
4 −
x
2
− 2 < ⇒
2 −
x
2
< ⇒
x
2
− 2 < (using the fact that |a − b| = |b − a|) ⇒
1
2
|x − 4| < ⇒
|x − 4| < 2 .
So, we choose δ = 2 . We can verify this choice of δ works because
0 < |x − 4| < δ = 2 ⇒
1
2
|x − 4| < ⇒
x
2
− 2 < ⇒
2 −
x
2
< ⇒
4 −
x
2
− 2 < .
This proves
lim
x→4
4 −
x
2
= 2.
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6. Example.
Prove that
lim
x→5
x2
+ 4 = 29
Solution.
Let > 0 be given. We must find δ as function of such that
x2
+ 4 − 29 < whenever 0 < |x − 5| < δ.
x2
+ 4 − 29 < ⇒
x2
− 25 < ⇒
|(x − 5)(x + 5)| < ⇒
|x − 5||x + 5| < (using the fact that |ab| = |a||b|).
Consider an interval of radius 1 around c = 5. For each x ∈ (4, 6), |x + 5| < 11. So,
|x − 5||x + 5| < 11|x − 5| < ⇒
11|x − 5| < ⇒
|x − 5| <
11
.
Choose
δ =
11
.
|x − 5| < δ =
11
⇒
11|x − 5| < ⇒
|x + 5||x − 5| <
|x2
− 25| <
|x2
+ 4 − 29| < ⇒
|(x2
+ 4) − 29| <
This proves
lim
x→5
(x2
+ 4) = 25
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