90 Chapter 3. Limits
Example Let f(x)= x
|x|.
• Forx>0, we have|x|= xand so f(x)= x
x =1.
Hence we get lim
x→0+
x
|x| =1.
• Forx<0, we have|x|=−xand so f(x)= x
−x =−1.
Hence we get lim
x→0−
x
|x| =−1.
Therefore, lim
x→0 x
|x|does not exist.
-3 -2 -1 1 2 3
-1 -0.5
0.5 1
Figure 3.17
The following rules are useful to find limits of functions at a pointa. In Rules (La4), (La5), (La5s) and (La6), f andgare functions that are defined on the left-side and right-side ofa. Some of the rules are similar to that for limits of functions at infinity.
Rules for Limits of Functions at a Point (La1) lim
x→ak=k (wherea∈Randkis a constant) (La2) lim
x→axn=an (wherea∈Randnis a positive integer) (La20) lim
x→a
√n
x= √n
a (wherea∈Randnis an odd positive integer) limx→a
√n
x= √n
a (where 0<a∈Randnis an even positive integer) (La3) lim
x→abx =ba (wherea∈Randbis a positive real number) (La4) lim
x→a
f(x)±g(x)
= lim
x→af(x)±lim
x→ag(x)
The result is valid for sum and difference of finitely many functions.
(La5) lim
x→a
f(x)·g(x)
= lim
x→af(x)·lim
x→ag(x)
The result is valid for product of finitely many functions.
(La5s) lim
x→a
k·g(x)
=k·lim
x→ag(x) (wherekis a constant) (La6) lim
x→a
f(x) g(x) =
x→alimf(x)
limx→ag(x) provided that lim
x→ag(x),0.
Example Find lim
x→4(1+x2), if it exists.
Solution lim
x→4(1+x2) = lim
x→41+lim
x→4x2 Rule (La4)
= 1+42 Rules (La1) and (La2)
= 17
Recall that a polynomial functionpis a function that can be written in the following form p(x)=cnxn+cn−1xn−1+· · ·+c1x+c0,
3.5. Two-sided Limits 91
wherec0,c1, . . . ,cnare constants. Using the method in the above example, we can prove the following theorem which means that the limit of a polynomial function at any real number can be found by substitution.
Theorem 3.5.1 Letp(x)be a polynomial and letabe a real number. Then we have limx→ap(x)= p(a).
Example Find lim
x→2
x−1
x2+x−2, if it exists.
Solution lim
x→2
x−1 x2+x−2 =
x→2lim(x−1)
limx→2(x2+x−2) Rule (La6)
= 2−1
22+2−2 Theorem 3.5.1
= 1
4
Recall that a rational functionris a function that can be written in the form r(x)= p(x)
q(x),
where pandqare polynomial functions. Using the method in the above example, we can prove the following theorem which means that the limit of a rational function at anya belonging to its domain can be found by substitution.
Theorem 3.5.2 Let p(x)andq(x)be polynomials and letabe a real number. Suppose thatq(a) ,0. Then we have
limx→a
p(x)
q(x) = p(a) q(a). Example Find lim
x→1
x−1
x2+x−2, if it exists.
ExplanationThe rational function f(x) = x−1
x2+x−2 is undefined atx = 1. This means that 1 does not belong to the domain of f and so we can’t apply Theorem 3.5.2. If we substitute x = 1 into the numerator and denominator, we get 0
0. We say that the limit is in theindeterminate form 0
0.
To find the limit, we replace f by a functiongwhich coincides with f on the left-side and right-side of 1 such that the limit ofgat 1 can be found by substitution (see the following figures).
-1 1 2 3 4 5
0.2 0.4 0.6 0.8 1
-1 1 2 3 4 5
0.2 0.4 0.6 0.8 1
y= f(x) y=g(x)
Figure 3.18(a) Figure 3.18(b)
92 Chapter 3. Limits
The functiongis given byg(x)= 1
x+2. It can be found by simplifying the expression defining f: f(x)= x−1
(x−1)(x+2) =g(x) for allx∈R\ {1,−2}.
Solution lim
x→1
x−1
x2+x−2 = lim
x→1
x−1 (x−1)(x+2)
= lim
x→1
1
x+2 Simplify expression
= 1
1+2 Theorem 3.5.2
= 1
3
Example Find lim
x→1
x+1
x2+x−2, if it exists.
ExplanationThe rational function f(x) = x+1
x2+x−2 is undefined at x = 1. Thus we can’t use Theorem 3.5.2.
If we put x = 1 into the numerator and denominator, we get 2
0. Limits (left-side and right-side) of the form
non-zero number
0 are∞or−∞. See the solution below for more details.
Solution lim
x→1
x+1
x2+x−2 does not exist. This is because ifxis close to 1, the numerator is close to 2 whereas the denominator is close to 0 and so the fraction is very large in magnitude (may be positive or
negative).
Remark Indeed, the denominator is x2 + x − 2 = (x− 1)(x +2).
Therefore, if x is close to and greater than 1, the denominator is small positive. Hence we have lim
x→1+
x+1
x2+x−2 = ∞. Similarly we have lim
x→1−
x+1
x2+x−2 =−∞.
-1 1 2 3
-20 -10 10 20
Figure 3.19
Example Let f(x)= x2+3. Find lim
h→0
f(x+h)− f(x)
h .
ExplanationThe expression f(x+h)−f(x)
h involves two variablesxandh. However, the question asks for lim (limit of the expression as happroaches 0). This implies that xis considered as a constant. In this way, theh→0
expression f(x+h)−f(x)
h is considered as a function ofh, defined for allh,0. The limit is in the indeterminate form 0
0 because if we puth=0 in the expression, the numerator and denominator are both 0. To find the limit, we simplify the expression so that the troublesome factorhin the denominator is canceled.
Solution f(x+h)− f(x)
h = (x+h)2+3
−(x2+3) h
= (x2+2xh+h2+3)−(x2+3) h
= 2xh+h2 h
= h(2x+h) h
= 2x+h
3.5. Two-sided Limits 93
Since f(x+h)− f(x)
h and 2x+h(considered as functions ofh) are equal on the left-side and right-side of h=0, it follows that
h→0lim
f(x+h)− f(x)
h = lim
h→0(2x+h)
= 2x+0 Theorem 3.5.1
= 2x
Remark The expression f(x+h)−f(x)
h is called a difference quotient. Limits of difference quotients will be discussed in detail in Chapter 4.
Summary for Limits In this chapter, we have introduced the following types of limits:
n→∞liman, lim
x→∞f(x), lim
x→−∞f(x), lim
x→a−f(x), lim
x→a+f(x), lim
x→af(x).
Since the definitions for lim
n→∞an and lim
x→∞f(x) are similar, we will omit limits of sequences in the following discussion. Note that for functions, the five types of limits take the form
x→lim f(x) wherecan be∞,−∞,a−,a+ora. The notation
x→lim f(x)= L, whereLis a real number, means the following
f(x) is arbitrarily close toLifxis “sufficiently close to (and different from)”, which, in short, is written as
f(x) is close toLifxis “close to (and different from)”, where f(x)is close to Lhas the usual meaning and
• x is “close to (and different from)”∞meansxis large;
• x is “close to (and different from)”−∞meansxis large negative;
• x is “close to (and different from)” a−meansxis close to and less thana;
• x is “close to (and different from)” a+meansxis close to and greater thana;
• x is “close to (and different from)” ahas the usual meaning.
If we cannot find a real numberLsuch that lim
x→f(x) = L, then we say that lim
x→f(x) does not exist. There are several possibilities for this. We may have
x→lim f(x)=∞ or lim
x→f(x)=−∞ or other behavior such as oscillation.
Similar to the above discussion, the notation lim
x→f(x)=∞means that f(x) is “close to”∞ifxis “close to (and different from)”, where“close to”∞means large.
94 Chapter 3. Limits
Exercise 3.5
1. For each of the following, find the limit, if it exists.
(a) lim
x→2(x2+3x−4) (b) lim
x→7(x2−5x−8)3 (c) lim
x→0
3x−5 2x+7
2
(d) lim
x→1 x√
x2+1 x+1
(e) lim
x→−225+4x (f) lim
x→−2 x2−4 x2+x−2
(g) lim
x→−2 x2−4
x2−x+2 (h) lim
x→6
x2−6x x2−5x−6
(i) lim
x→5 x2+25
x−5 (j) lim
x→3 x−3 x2+9
(k) lim
x→4
√x−2
x−4 (l) lim
x→2 x−5 2−√
x−1
2. For each of the following f, find lim
h→0
f(x+h)−f(x)
h .
(a) f(x)=4x−13 (b) f(x)= x3 (c) f(x)= 1
x (d) f(x)=√
x