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MATH 1910 - Limits Numerically and Graphically
Introduction to Limits
 The concept of a limit is our doorway to calculus. This lecture will explain what the limit of a
function is and how we can find such a limit. Be sure you understand function notation at this point,
it will be used throughout the remainder of the course.
 Consider the function
4
f x  x − 16
x−2
Note that the domain of f is x | x ≠ 2. What does the graph look like near x  2? We can certainly
graph the function with our graphing calculator & see what happens. Before we do this, though, let’s
look at the value (output) of f for values of x close to 2. These can be seen in the following table.
x
1.99
1.999
1.9999 2.0001
2.001
2.01
fx ≈ 31.761 31.976 31.998 32.002 32.024 32.241
Note that as x approaches (gets close to) 2, the value of f x seems to be approaching 32. We say
4
"The limit of fx  x − 16 as x approaches 2 is 32"
x−2
and write
4
lim f x  32 OR lim x − 16  32
x→2
x→2 x − 2
Graph the function over the intervals 0 ≤ x ≤ 4 and 0 ≤ y ≤ 40. Describe what is happening to the
graph of f as x approaches 2. Note how the graph shows the same behavior as the table above
describes.
 The following is an intuitive definition for the limit of a function: If f x gets arbitrarily close to a
real number L as x approaches (gets close to) a, then
f x  L OR f x  L as x  a
lim
x→a
We say this as "the limit of fx as x approaches a is L".
 The phrase "gets arbitrarily close to" basically means as close as we like.
 If f x does NOT get arbitrarily close to a real number L, we say that the limit does not exist. I
will write DNE from now on if the limit does not exist.
 Note that in the previous example f 2 does not exist (is undefined), but lim f x DOES exist.
x→2
Hence, for a limit to exist at a, the function does not have to be defined at a.
Finding Limits Numerically & Graphically
 When finding limits numerically we will basically construct a table of values as we did in the
example above. When finding limits graphically we will look at the graph of the function to estimate
limits. Here are some examples:
1. Estimate numerically lim gx if
x→9
x −3
x−9
We construct a table of values for gx for values of x close to 9.
gx 
x
8.9
8.99
8.999
9.001
9.01
9.1
gx ≈ 0.16713 0.16671 0.16667 0.16666 0.16662 0.16621
It appears that as x approaches 9 that gx is getting closer to 0.16666... or 0. 1 6 
1
6
(see below).
1
Hence, it appears that
lim gx  1
x→9
6
Side Note: Do you know how to convert from a non-terminating but repeating decimal expansion like
0. 1 6 to its equivalent fraction? Here’s one way:
Let n  0. 1 6 . Then 10n  1. 6 and 100n  16. 6 . Thus
90n  100n − 10n  16. 6 − 1. 6  15
Thus
90n  15  n  15  1
90
6
sin
x
2. Construct a table of values for f x  x for x close to zero to estimate
x
lim sin
x
x→0
What mode should we be in? (radian or degree?)
Here is such a table:
x
0. 1
0. 01
0. 001
sin x ≈ 0.99833417 0.9999833 0.9999998
x
Thus it appears that
x
lim sin
x 1
x→0
This is an important limit we will see again. Look at the graph of sinx x over the intervals − ≤ x ≤ 
& −2 ≤ y ≤ 2 to confirm the numerical approach.
3. An example of a limit that does not exist (DNE).
Consider the function f x  sin 1x . Note that the domain of f is all real numbers except 0. What can
we say about
lim sin 1x
x→0
When we try to graph this function for values of x near zero, our graphing calculator has problems.
Graph f over the intervals −2 ≤ y ≤ 2 and (1) −3 ≤ x ≤ 3, (2) −1 ≤ x ≤ 1, (3) −0. 1 ≤ x ≤ 0. 1, and
finally (4) −0. 01 ≤ x ≤ 0. 01. What do you observe? What is happening?
Recall that
1  4n
sin t  1 if t    2n    4n 
for n  0, 1, 2, 3, . . .
2
2
2
2
and
3  4n
for n  0, 1, 2, 3, . . .
sin t  −1 if t  3  2n  3  4n 
2
2
2
2
So
1  4n
2
or when x 
for n  0, 1, 2, 3, . . .
sin 1x  1 when 1x 
2
1  4n
Thus f x  1 when
2 , 2 , 2 , 2 ,...,
2
x 
,...
5 9 13
1  4n
Note that as n → , x → 0. Or saying it another way, between any positive number x and zero, there
are an INFINITE number of times when f x  1. In a similar manner,
2
3  4n
2
sin 1x  −1 when 1x 
or when x 
for n  0, 1, 2, 3, . . .
2
3  4n
Thus f x  −1 when
2
x  2 , 2 , 2 , 2 ,...,
,...
3 7 11 15
3  4n
Note that as n → , x → 0. Or saying it another way, between any positive number x and zero, there
are an INFINITE number of times when f x  −1.
Thus we see that as x gets close to zero, f x begins to wildly oscillate between −1 and 1. In essence,
f x can never "settle down" and approach any one limit. Thus lim sin 1x DNE.
x→0
4. When Technology Fails.
We saw in the last example that our graphing calculator had troubles graphing the function for values
of x close to zero. This example shows another type error we can run into.
t4  1 − 1
If gt 
, estimate numerically the following limit
t4
lim gt
t→0
We proceed by constructing a table of values for x close to zero
t
0. 1
0. 01 0. 001 0. 0001
gt 0.49999
0.5
0
0
Up to x  0. 01, we may guess that the limit appears to be approaching 12 . But, if we get closer to
zero, we see the limit appears to be zero. What is going on?
The problem is that when your graphing calculator evaluates t 4  1 for small values of t, the result
is very close to 1. In fact, if you plug in t  0. 001 in the TI-84 the result is zero. This is due to the
limitations of the graphing calculator & the number of digits the calculator is able to carry (Graph g
with 0 ≤ y ≤ 1 over (1) −4 ≤ x ≤ 4 and the (2) −0. 01 ≤ x ≤ 0. 01). The value of this limit is 12
which will be able to show later.
One-Sided Limits
 Consider the function
f x 
1 if x  2
3 if x  2
The graph of f is shown below
4
3
2
1
-2
2
4
6
Note that as x approaches 2 from the left (or from the negative side or from below) fx approaches 1
(it is always 1 for x  2). But as x approaches 2 from the right (or from the positive side or from
above) fx approaches 3. Since we do not approach any ONE value from both "sides", lim fx DNE.
x→2
When fx approaches 1 as x approaches 2 from the left we write
3
lim f x  1
x → 2−
and we say "the limit of fx as x approaches 2 from the left is 1" or "the left-hand limit of fx as x
approaches 2 is 1". Similarly, When fx approaches 3 as x approaches 2 from the right we write
lim f x  3
x → 2
and we say "the limit of fx as x approaches 2 from the right is 3" or "the right-hand limit of fx as x
approaches 2 is 3".
 With one-sided limits we have the following useful theorem
lim
f x  L if and only if xlim
f x  L  xlim
f x
x→a
→a−
→a
fx  L if and only if both one-sided limits exist and are both equal to L.
That is, lim
x→a
 Consider the function
−1 if x  0
|x|
x 
1
if x  0
The graph of the function is shown below
1
-4
-2
2
4
-1
|x|
|x|
Note that lim x  −1, but lim x  1. Both one-sided limits exist, but they are not equal. Thus
x→0−
x→0
|x|
lim x DNE.
x→0
 Consider the function hx whose graph is shown below. Find the following limits (if they exist)
(a) lim hx (b) lim hx (c) lim hx (d) lim hx (e) lim hx (f) lim hx
x → −2 −
x → −2 
x → −2
x→1−
x→1
x→1
4
2
-5
-4
-3
-2
-1
1
2
3
4
5
-2
-4
Infinite Limits
Let’s try to find the limit lim 12 We proceed numerically, constructing a table of values for x close to
x→0 x
zero.
x
1
x2
0. 1 0. 01
0. 001
0. 0001
100 10,000 1,000,000 100,000,000
4
As x gets closer to zero, the value of 12 continues to get bigger and bigger. It does NOT approach any
x
finite number. Thus, we approach no FINITE limit. To indicate this kind of behavior we introduce the
notation
lim 12  
x→0 x
and say that f has an infinite limit. Note that this does NOT mean  is a number (which it is not). It
simply expresses the idea that the value of the function gets arbitrarily large (as large as we want) as x
gets close to 0. When we see this expression, we say "the limit is infinity" or "the function increases
without bound".
If a function f gets arbitrarily large BUT NEGATIVE as x approaches a, we write
lim
fx  −
x→a
We can say similar statements with one-sided limits.
Examples:
1. For the function g shown below find the following limits or write DNE.
lim gx
lim gx
lim gx
x → −2 −
x → −2 
lim gx
x → −2
lim gx
x→3−
lim gx
x→3
x→3
10
5
-4
-3
-2
-1
1
2
3
4
-5
-10
2. Show that lim ln x  −.
x→0
5
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