1S11: Calculus for students in Science
Dr. Vladimir Dotsenko
TCD
Lecture 10
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
1/1
Limits at infinity
A function f is said to have the limit L as x tends to +∞ if the values
f (x) get as close as we like to L as x increases without bound. In this
case, one writes
lim f (x) = L.
x→+∞
Similarly, a function f is said to have the limit L as x tends to −∞ if the
values f (x) get as close as we like to L as x decreases without bound. In
this case, one writes
lim f (x) = L.
x→−∞
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
2/1
Limits at infinity
For example, let us consider the function f (x) = 1/x:
y
x
In this case, we have
lim f (x) = 0 and
x→+∞
Dr. Vladimir Dotsenko (TCD)
lim f (x) = 0.
x→−∞
1S11: Calculus for students in Science
Lecture 10
3/1
Limits at infinity
Now, let us consider the function f (x) = 1/x 2 :
y
x
In this case, we also have
lim f (x) = 0 and
x→+∞
Dr. Vladimir Dotsenko (TCD)
lim f (x) = 0.
x→−∞
1S11: Calculus for students in Science
Lecture 10
4/1
Limits at infinity
In terms of limits at infinity we can make the notion of a horisontal
asymptote more precise:
The graph of a function f has the line y = L as a horisontal asymptote if
at least one of the two following situations occur:
lim f (x) = L, lim f (x) = L.
x→+∞
x→−∞
E.g., both for f (x) = 1/x and f (x) = 1/x 2 , both of the formulas apply
when L = 0.
Exercise. Sketch examples of graphs for which exactly one of those
situations occurs.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
5/1
Infinite limits at infinity
A function f is said to have the limit +∞ as x tends to +∞ (or −∞) if
the values f (x) all increase without bound as x increases without bound
(decreases without bound). In this case, one writes
lim f (x) = +∞
x→+∞
( lim f (x) = +∞).
x→−∞
A function f is said to have the limit −∞ as x tends to +∞ (or −∞) if
the values f (x) all decrease without bound as x increases without bound
(decreases without bound). In this case, one writes
lim f (x) = −∞
x→+∞
Dr. Vladimir Dotsenko (TCD)
( lim f (x) = −∞).
x→−∞
1S11: Calculus for students in Science
Lecture 10
6/1
Infinite limits at infinity
For example, let us consider the function f (x) = x 2 :
y
x
In this case, we have
lim f (x) = +∞ and
x→+∞
Dr. Vladimir Dotsenko (TCD)
lim f (x) = +∞.
x→−∞
1S11: Calculus for students in Science
Lecture 10
7/1
Infinite limits at infinity
Now, let us consider the function f (x) = x 3 :
y
x
In this case, we have
lim f (x) = +∞ and
x→+∞
Dr. Vladimir Dotsenko (TCD)
lim f (x) = −∞.
x→−∞
1S11: Calculus for students in Science
Lecture 10
8/1
Limits of polynomials at infinity
Theorem. Suppose that f (x) = a0 + a1 x + · · · + an x n , where an 6= 0.
Then
lim f (x) = lim (an x n ),
x→+∞
x→+∞
lim f (x) = lim (an x n ).
x→−∞
x→−∞
Proof. We have
n
f (x) = a0 + a1 x + · · · + an x = an x
n
a1
an−1
a0
+1 ,
+
+ ··· +
an x n
an x n−1
an x
and the expression in the brackets clearly has the limit 1 as x → +∞ or
x → −∞, since all terms except for 1 have the limit 0.
Informally, this theorem says that the limiting behaviour at infinity of a
polynomial exactly matches the behaviour of its highest degree term.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
9/1
Limits of rational functions at infinity
Theorem. Suppose that f (x) =
a0 +a1 x+···+an x n
b0 +b1 x+···+bm x m ,
where an , bm 6= 0. Then
an x n
,
x→+∞
x→+∞ bm x m
an x n
.
lim f (x) = lim
x→−∞
x→−∞ bm x m
lim f (x) = lim
Proof. We have
f (x) =
an x n
a0 + a1 x + · · · + an x n
=
·
m
b0 + b1 x + · · · + bm x
bm x m
a0
an x n
b0
bm x m
+
+
a1
+ · · · + aan−1
+1
an x n−1
nx
,
b
b1
m−1
+
1
+
·
·
·
+
m−1
bm x
bm x
and both the numerator and the denominator of the second fraction
clearly have the limit 1 as x → +∞ or x → −∞, since all terms except
for 1 have the limit 0.
Informally, this theorem says that the limiting behaviour at infinity of a
rational function exactly matches the behaviour of the ratio of highest
degree terms in the numerator and the denominator.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
10 / 1
Limits of rational functions at infinity:
examples
Example 1. We have
3x + 5
3x 1 +
= lim
·
x→+∞ 6x − 8
x→+∞ 6x 1 −
lim
5
3x
8
6x
3x
1
= .
x→+∞ 6x
2
= lim
Example 2. We have
1
4x 2 1 − 4x
2
4x 2
4x 2 − x
=
lim
·
= lim
= 0.
=
lim
5
3
x→−∞ 2x 3 1 −
x→−∞
x→−∞
x→−∞ 2x 3 − 5
2x
x
3
2x
lim
Example 3. We have
2
+ 5x1 3
5x 3 − 2x 2 + 1
5x 3 1 − 5x
5x 2
= lim
·
= −∞.
=
lim
1
x→+∞
x→+∞ −3x
x→+∞ −3
1 − 3x
1 − 3x
lim
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
11 / 1
One example from Lecture 5
y
y=
x 2 +2x
x 2 −1
x
2
2
x +2x
= (x−1)(x+1)
has vertical asymptotes x = 1
The graph of f (x) = xx 2+2x
−1
and x = −1, and a horisontal asymptote y = 1. Now we know which
limits control those asymptotes:
lim f (x) = +∞, lim f (x) = −∞, lim + f (x) = +∞, lim f (x) = −∞
x→1+
x→1−
x→−1
x→−1−
for the vertical asymptotes, and limx→+∞ f (x) = 1, limx→−∞ f (x) = 1 for
the horisontal asymptote.
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
12 / 1
Another example from Lecture 5
y
y=
2x 2 −2
x 2 −2x−3
bc
x
2
−2
The graph of f (x) = x 22x−2x−3
= 2(x−1)(x+1)
(x+1)(x−3) has a vertical asymptote
x = 3, a horisontal asymptote y = 2, and also the point (−1, 1) which it
approaches both on the left and on the right but does not touch. This
corresponds to the existence of the limits
lim f (x) = +∞, lim f (x) = −∞,
x→3+
x→3−
lim f (x) = 2, lim f (x) = 2,
x→+∞
x→−∞
lim f (x) = 1.
x→−1
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
13 / 1
More difficult limits at infinity
For ratios involving square roots, the same method we used before is
useful:
q
√
x 2 (1 + x22 )
2
1
x +2
|x|
= lim
= ,
lim
= lim
x→+∞ 3x(1 − 2 )
x→+∞ 3x − 6
x→+∞ 3x
3
x
q
√
x 2 (1 + x22 )
2
|x|
x +2
1
lim
= lim
= lim
=− .
2
x→−∞ 3x − 6
x→−∞ 3x
x→−∞ 3x(1 − )
3
x
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
14 / 1
More difficult
limits
at infinity
p
p
p
For differences f (x) − g (x), or simply
2 −b 2
apply the formula a − b = aa+b
:
f (x) − h(x), it is useful to
√
p
( x 6 + 5)2 − (x 3 )2
3
6
√
lim ( x + 5 − x ) = lim
=
x→+∞
x→+∞
x6 + 5 + x3
= lim √
x→+∞
5
= 0,
x6 + 5 + x3
√
p
x 6 + 5x 3 )2 − (x 3 )2
(
3
√
=
lim ( x 6 + 5x 3 − x ) = lim
x→+∞
x→+∞
x 6 + 5x 3 + x 3
5x 3
= lim √
,
x→+∞
x 6 + 5x 3 + x 3
and noticing that for large positive
x we can
q
use our previous method and
√
5
3
3
6
3
1 + x 3 + 1 , we finally conclude that
write x + 5x + x = x
√
5
limx→+∞( x 6 + 5x 3 − x 3 ) = 2 .
Dr. Vladimir Dotsenko (TCD)
1S11: Calculus for students in Science
Lecture 10
15 / 1