80 Chapter 3. Limits
∗4. Suppose (an) is a sequence such that 0< anfor allnanda1 > a2 >a3 > · · ·. Does lim
n→∞anexist? What can you tell about the limit?
3.3. Limits of Functions at Infinity 81
The following rules for limits of functions at infinity are similar to that for limits of sequences. In (4), (5), (5s) and (6), f andgare functions such that f(x) andg(x) are defined for sufficiently largex.
Rules for Limits of Functions at Infinity (L1) lim
x→∞k=k (wherekis a constant) (L2) lim
x→∞
1
xp =0 (wherepis a positive constant) (L3) lim
x→∞
1
bx =0 (wherebis a constant greater than 1) (L4) lim
x→∞
f(x)±g(x)
= lim
x→∞f(x)± lim
x→∞g(x)
The result is valid for sum and difference of finitely many functions.
(L5) lim
x→∞
f(x)·g(x)
= lim
x→∞f(x)· lim
x→∞g(x)
The result is valid for product of finitely many functions.
(L5s) lim
x→∞
k·g(x)
=k· lim
x→∞g(x) (L6) lim
x→∞
f(x) g(x) =
x→∞lim f(x)
x→∞lim g(x) provided that lim
x→∞g(x),0.
To consider limits of functions at infinity, we should first check the domains of the functions. For example, if f(x)= √
1−x, the domain of f is{x∈R: 1−x ≥0}=(−∞,1]; it is meaningless to talk about limit of f at infinity. In the next example, the domain of the function 1− 2
x3 isR\ {0}; the function is defined for largexand hence we may consider its limit at infinity (whether the limit exists; and if exists, find the value).
Example Find lim
x→∞ 1− 2 x3
!
, if it exists.
Solution lim
x→∞ 1− 2 x3
!
= lim
x→∞1− lim
x→∞ 2· 1 x3
!
Rule (L4), rewrite 2nd term
= 1−2· lim
x→∞
1
x3 Rules (L1) and (L5s)
= 1−2·0 Rule (L2)
= 1
Example Find lim
x→∞ 2−x+3
, if it exists.
Solution lim
x→∞(2−x+3) = lim
x→∞2−x+ lim
x→∞3 Rule (L4)
= lim
x→∞
1
2x +3 Rewrite first term and Rule (L1)
= 0+3 Rule (L3)
= 3
82 Chapter 3. Limits
Example Find lim
x→∞
x2+1
3x3−4x+5, if it exists.
ExplanationBecause limits of the numerator and denominator do not exist, we can’t apply Rule (6). The first step is to divide the numerator and denominator by x3 so that the limits at infinity of the new numerator and denominator exist.
Solution
x→∞lim
x2+1
3x3−4x+5 = lim
x→∞
x2+1 x3 3x3−4x+5
x3
Divide numerator and denominator byx3
=
x→∞lim 1
x + 1
x3
x→∞lim
3− 4
x2 + 5
x3
Rule (L6), rewrite numerator and denominator
=
x→∞lim
1 x + lim
x→∞
1 x3 x→∞lim3− lim
x→∞
4 x2 + lim
x→∞
5 x3
Rule (L4)
= 0+0
3−0+0 Rules (L1), (L2) and (L5s)
= 0
The next example is similar to the last one. To find limits at infinity for rational functions, we can divide the numerator and denominator by a suitable power ofx.
Example Find lim
x→∞
x3+1
3x3−4x+5, if it exists.
Solution
x→∞lim
x3+1
3x3−4x+5 = lim
x→∞
x3+1 x3 3x3−4x+5
x3
Divide numerator and denominator byx3
=
x→∞lim
1+ 1
x3
x→∞lim
3− 4
x2 + 5
x3
Rule (L6), rewrite numerator and denominator
=
x→∞lim 1+ lim
x→∞
1 x3 x→∞lim3− lim
x→∞
4 x2 + 5
x3
Rule (L4)
= 1+0
3−0+0 Rules (L1), (L2) and (L5s)
= 1
3
To find limits at infinity for rational functions, we can also use the following shortcut.
3.3. Limits of Functions at Infinity 83
Leading Terms RuleLet f(x)=anxn+an−1xn−1+· · ·+a1x+a0andg(x)=bmxm+bm−1xm−1+· · ·+b1x+b0, wherean ,0 andbm,0. Then we have
x→∞lim f(x) g(x) = lim
x→∞
anxn+an−1xn−1+· · ·+a1x+a0
bmxm+bm−1xm−1+· · ·+b1x+b0 = lim
x→∞
anxn bmxm Proof The idea is to extract factoranxnin the numerator andbmxmin the denominator. Putting
ϕ(x)= 1+ aan−1
n · 1x + aan−2
n · x12 +· · ·+ aa1
n · xn−11 + aa0
n · x1n
1+bbm−1m · 1x + bbm−2m · x12 +· · ·+ bbm1 · xm−11 + bbm0 · x1m
we have f(x)
g(x) = anxn
bmxm ·ϕ(x). It is straightforward to check that lim
x→∞ϕ(x) = 1. Hence by Rule (5), we obtain
the required result.
Remark
• (a) Ifn=m, the limit is an
bn. (b) Ifn<m, the limit is lim
x→∞
an
bm
· 1
xm−n
=0.
(c) Ifn > m, the limit is lim
x→∞
an
bm
·xn−m
which does not exist because as xincreases indefinitely, xn−m increases indefinitely.
• The Leading Terms Rule can also be applied to “functions similar to rational functions”, for example, for f(x)= x+2√
x+3 andg(x)=5x+6√
x+7, we have lim
x→∞
x+2√ x+3 5x+6√
x+7 = lim
x→∞
x 5x
• The Leading Terms Rule can’t be applied to limits of rational functions at a point: lim
x→a f(x)
g(x), wherea∈R.
Below we re-do the last two examples using the Leading Terms Rule.
Example lim
x→∞
x2+1
3x3−4x+5 = lim
x→∞
x2
3x3 Leading Terms Rule
= lim
x→∞
1 3 ·1
x
!
Simplify and rewrite expression
= 1
3 ·0 Rules (L2) and (L5s)
= 0 Example lim
x→∞
x3+1
3x3−4x+5 = lim
x→∞
x3
3x3 Leading Terms Rule
= lim
x→∞
1
3 Simplify expression
= 1
3 Rule (L1)
84 Chapter 3. Limits
In the next example, the function can be considered as a product or a quotient of two functions. However, we can’t apply Rule (5) or (6) because limit at infinity of one of the functions does not exist. To find the limit, we need the following result.
Sandwich TheoremLet f,gandhbe functions such that f(x),g(x)andh(x)are defined for sufficiently large x. Suppose that f(x) ≤ g(x) ≤ h(x)if xis sufficiently large and that both lim
x→∞f(x)and lim
x→∞h(x)exist and are equal (with common limit denoted byL). Then we have lim
x→∞g(x)=L.
RemarkThe condition “f(x) ≤ g(x) ≤ h(x) if x is sufficiently large” means that there is a real numberrsuch that the inequalities are true for all x>r.
Example Find lim
x→∞
sinx
x , if it exists.
Explanation The given function can be written as a product of two functions: sinx and 1
x. For the second function, its limit at infinity is 0. However, for the first function, its limit at infinity does not exist. Thus we can’t apply Rule (5).
Solution Since−1≤sinx≤1 for all real numbersx, it follows that
−1
x ≤ sinx x ≤ 1
x for allx>0.
Note that lim
x→∞
−1 x = lim
x→∞
1
x =0. Thus by the Sandwich Theorem, we have lim
x→∞
sinx
x =0.
Example Find lim
x→∞(1+logx), if it exists.
RemarkSince lim
x→∞logxdoes not exist, we can’t apply Rule (L4).
Solution lim
x→∞(1+logx) does not exist. This is because ifxincreases without bound, so does 1+logx.
Infinite Limits
In the last example, although limit does not exist, we know that ifxincreases indefinitely, so does (1+logx).
In the limit notation lim
x→∞, the symbolx → ∞indicates that “x increases indefinitely”, or “x approaches∞”.
Using the same idea, we also write 1+logx → ∞which indicates that the value increases indefinitely (as x increases indefinitely). Puttingy=1+logx, we writey→ ∞asx→ ∞. Concerning the graph ofy= f(x) in the coordinate plane, x→ ∞means thatxgoes to the right indefinitely, approaching the point∞(an imaginary point on the right) andy → ∞means thatygoes up indefinitely, approaching the point∞(an imaginary point at the top).
Notation Let f be a function such that f(x) is defined for sufficiently largex. Suppose that (∗) f(x) is arbitrarily large ifxis sufficiently large.
Then we write lim
x→∞f(x)=∞.
Remark
• Because∞is not a real number, lim
x→∞f(x) = ∞does not mean the limit exists. In fact, it indicates that the limit does not exist and explains why it does not exist.
3.3. Limits of Functions at Infinity 85
• Instead of lim
x→∞f(x)=∞, we also write f(x)→ ∞asx→ ∞.
• For simplicity, instead of saying Condition (∗), we will say (∗∗) f(x) is large ifxis large.
• Similar to lim
x→∞f(x)=∞, we also have lim
x→∞f(x)=−∞which means that (∗∗) f(x) is large negative ifxis large.
Example (a) lim
x→∞(1+x2)=∞ (limit does not exist) (b) lim
x→∞(1−x2)=−∞ (limit does not exist)
2 4 6 8 10
20 40 60 80 100
Figure 3.6
2 4 6 8 10
-100 -80 -60 -40 -20
Figure 3.7
Example lim
x→∞
1+x2
1+x = lim
x→∞
x2
x Leading Terms Rule
= lim
x→∞x
= ∞ (limit does not exist)
2 4 6 8 10
2 4 6 8
Figure 3.8
Limits at negative infinity
Similar to limits at infinity, we may consider limits at negative infinity provided that f(x) is defined for x sufficiently large negative. Readers can figure out the meaning of the following notations:
• lim
x→−∞f(x)=LwhereLis a real number;
• lim
x→−∞f(x)=∞;
• lim
x→−∞f(x)=−∞.
Example (1) lim
x→−∞
1 x =0 (2) lim
x→−∞x3=−∞ (limit does not exist) (3) lim
x→−∞(1−x3)=∞ (limit does not exist)
FAQ Can we perform addition, multiplication etc. with∞or−∞?
Answer Yes and no. For example,
• 1+∞=∞,
86 Chapter 3. Limits
• 2· ∞=∞.
However,
• ∞ − ∞is undefined,
• 0· ∞is undefined.
Be careful when you perform such operations.
Exercise 3.3
1. For each of the following, find the limit if it exists.
(a) lim
x→∞
√1
x−1 (b) lim
x→∞(15−16x−3) (c) lim
x→∞5−x (d) lim
x→∞
√x (e) lim
x→∞
x2+9
x3+1 (f) lim
x→∞
x2+9 x2+1
(g) lim
x→∞
x2+9
x+1 (h) lim
x→∞
|x|
x
(i) lim
x→−∞
|x|
x (j) lim
x→−∞xsinx
2. The concentrationCof a drug in a patient’s bloodstreamthours after it was injected is given by C(t)= 0.15t
t2+3. (a) Find lim
t→∞C(t).
(b) Interpret the result in (a).
3. The populationPof a certain small towntyears from now is predicted to beP(t)=35000+ 10000
(t+2)2. (a) Find the population in the long run.
(b) Use computer to sketch the graph ofP. What can you tell from the graph?
∗4. For each of the following functions f, use computer to find f(x) for large x. Guess whether lim
x→∞f(x) exists or not. If the limit exists, what is the limit?
(a) f(x)=√
x+1−√ x (b) f(x)=√
x2+x−x (c) f(x)= x99
2x