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2.4 The Precise Definition of a Limit
Reminders/Remarks: |x − 4| < 3 means that the distance between x and 4 is less than 3. In other words, x
lies strictly between 1 and 7. So, |x − a| < δ means that the distance between x and a is less than δ, that
is, x lies within δ of a.
Note: The inequality 0 < |x − a| < δ means the distance between x and a is less than δ, but x 6= a.
Similarly, if we write |f (x) − L| < 2, this means that f (x) lies within 2 of L.
Consider the function f (x) = 2x + 4. We know lim (2x + 4) = 10. We said that the limit means we can
x→3
make f (x) as close to 10 as we want by getting x closer and closer to 3.
How close to 3 does x need to be so that f (x) lies within 1 of the limit value 10, i.e. |f (x) − 10| < 1 ?
How close to 3 does x need to be so that f (x) lies within 0.1 of the limit value 10, i.e. |f (x) − 10| < 0.1 ?
How close to 3 does x need to be so that f (x) lies within an arbitrary number ǫ of 10, i.e. |f (x) − 10| < ǫ?
This number that we’re finding is denoted as δ.
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ǫ, δ Definition of a Limit: Let f be a function defined on some open interval that contains the number a,
except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write
lim f (x) = L
x→a
if for every number ǫ > 0 there exists a number δ > 0 such that whenever 0 < |x−a| < δ, then |f (x)−L| < ǫ.
To prove a limit using the definition of the limit, there are two steps:
1. Do some scratchwork to determine a value for δ.
2. Show that this δ works.
Example: Prove using the definition of the limit that lim (2x + 4) = 10.
x→3
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Example: Prove using the definition of the limit that lim (3x + 7) = 1.
x→−2
Consider the function f (x) = x2 . We know lim x2 = 4.
x→2
Question: Given ǫ = 1, find a number δ so that if |x − 2| < δ, then |x2 − 4| < 1.
Question: Given ǫ = 0.1, find a number δ so that if |x − 2| < δ, then |x2 − 4| < 0.1.
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1
Consider the function f (x) = 2 . We know lim 2 = ∞. We said that this means we can make f (x) as
x→0 x
x
large as we want by getting x closer and closer to 0.
Definition: Left f be a function defined on some open interval that contains the number a, except possibly
at a itself. Then
lim f (x) = ∞
x→a
if for every number M > 0 there exists a number δ > 0 such that whenever 0 < |x − a| < δ, then f (x) > M .
Find a number δ so that if 0 < |x| < δ, then
1
> 100.
x2
Find a number δ so that if 0 < |x| < δ, then
1
> 10, 000.
x2
Find a number δ so that if 0 < |x| < δ, then
1
> M , where M is some arbitrary number.
x2
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2.5 Continuity
In Section 2.3 we saw that the limit as x approaches a can sometimes be found by evaluating the function
at a. If this is the case, then the function is continuous.
Definition: A function is continuous at a number a if
lim f (x) = f (a)
x→a
Otherwise, we say the function is discontinuous at a, or that there is a discontinuity at a.
In order for a function to be continuous at a number a:
(1) f (a) must be defined. – So a function will NOT be continuous anywhere it is undefined.
(2) lim f (x) must exist. (The left-handed and right-handed limits must both equal the same value.)
x→a
(3) lim f (x) = f (a)
x→a
Examples of discontinuities: Holes, vertical asymptotes, and jumps.
A “hole” in a graph is also referred to as a removable discontinuity because if we wanted to, we could
just redefine the function at that point to make it continuous. Removable discontinuities occur where the
limit exists at a (left and right limits are equal), but is not equal to f (a).
A vertical asymptote is referred to as an infinite discontinuity.
A jump in the graph is referred to as a jump discontinuity. Jumps occur where the limits from the left
and right exist, but are not equal.
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A function is continuous from the left at a number a if lim f (x) = f (a) and continuous from the
x→a−
right if lim+ f (x) = f (a). A function is continuous if and only if it is continuous from both the right and
the left.
x→a
Examples: Determine where the functions below are discontinuous. State the type of discontinuity and
explain why mathematically using limits. Is the function continuous from the left or right there?
(1) f (x) =
x2 − 25
x−5
(2) f (x) =
(x + 3)(x − 2)
(x − 2)3
(3) f (x) =
(
x2 − 4
x+1
if x ≤ −1
if x > −1
Fact: All polynomials are continuous everywhere!
Fact: A rational function is continuous wherever it is defined, i.e. where the denominator is not 0.
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(4) f (x) =


 2x − 1
6

 x2 − 9
(5) f (x) =

3x + 1




 x2 − 5
x−1

3 − 25


x


x−2
if x < 4
if x = 4
if x > 4
if x < −2
if − 2 ≤ x ≤ 3
if x > 3
What values of a, b, and c would make the following function continuous everywhere?


ax2 + bx + 1


 3x + 2a + b
f (x) =

c


 2
x − ax − b
if
if
if
if
x ≤ −1
−1<x<2
x=2
x>2
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If f and g are continuous at a and c is any constant, then the functions f + g, f − g, cf , f g, and
g(a) 6= 0) are all continuous functions.
f
g
(where
The Intermediate Value Theorem: Suppose f is continuous on the closed interval [a, b] and let N be any
number strictly between f (a) and f (b). Then there exists a number c in (a, b) such that f (c) = N .
Example: Show that the equation −x3 + 2x + 2 = 0 has a root (solution) on the interval (1, 2).
Example: If f (x) = x4 − x3 + 3x2 + 2, show that there is a number c so that f (c) = 3.
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2.6 Limits at Infinity; Horizontal Asymptotes
Up to this point, we have dealt with limits as x approaches some number a. Now, we examine limits as x
approaches ∞ or −∞. These are called limits at infinity.
Let f be a function defined on some interval (a, ∞). Then,
lim f (x) = L
x→∞
means that the values of f (x) can be made arbitrarily close to L by taking x sufficiently large.
Definition: If lim f (x) = L or lim f (x) = L, then f (x) has a horizontal asymptote at y = L.
x→∞
lim 1
x→∞ x
=
lim (x3 − x2 )
x→∞
x→−∞
lim 2
x→−∞ x
lim 32
x→±∞ x
=
lim (x3 − x2 )
x→−∞
Evaluate the following limits:
2x3 + x2 − 1
x→∞ 5x3 − 7x + 2
lim
2x3 + x2 − 1
x→−∞ 5x3 − 7x + 2
lim
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4x2 − 2x + 3
x→−∞
5 − 3x
4x2 − 2x + 3
x→∞
5 − 3x
lim
lim
x2 (x − 8)
x→∞ (x2 + 1)(x2 − 3)
x2 (x − 8)
x→∞ (x2 + 1)(x2 − 3)
lim
lim
lim
√
x→∞
lim
x→−∞
9x2 − 12
5x + 2
√
9x2 − 12
5x + 2
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lim
x→∞
lim
√
x→−∞
x2 − x − x
√
x2 + 5x + x
5 − 4x3
Find all horizontal asymptotes of the function f (x) = √
.
25x6 + 1
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