2.3 Calculating Limits Using the Limit Laws
2.3 Calculating Limits Using the Limit Laws
Calculating limits by testing values of x close to a is tedious. The following Theorem essentially says that any `nice' combination of functions has exactly the limit you'd expect.
Theorem.
Suppose
lim
xa
f (x) and
lim
xa
g(x)
both
exist
and
that
c
is
constant.
Then the following limits exist and may be computed.
1. xlima c = c
2.
lim
xa
x
=
a
3. xlima c f (x) = c xlima f (x)
4.
xlima[
f
(x)
?
g(x)]
=
lim
xa
f
(x)
?
lim
xa
g(x)
5.
xlima(
f
(x)g(x))
=
lim
xa
f
(x)
?
lim
xa
g(x)
6.
lim
xa
f (x) g(x)
=
lim
xa
f
(x)
lim
xa
g(x)
provided
lim
xa
g(x)
=
0
The Theorem also hold for one-sided limits and, with a little care,1 for infinite limits. For example,
if lim f (x) = -3 and lim g(x) = -, then
x2
x2
lim[ f (x) + g(x)] = -
x2
and lim f (x)g(x) =
x2
Corollary.
Suppose
that
p(x)
=
cn xn
+ cn-1xn-1
+ ? ? ? + c1x
+ c0
is
a
polynomial, then
lim
xa
p(x)
=
p(a).
Moreover, if r is any rational function, and a dom(r), then xlima r(x) = f (a).
Proof. Simply calculate:
lim
xa
p(x)
=
lim
xa
cn xn
+
?
?
?
+
c1
lim
xa
x
+
lim
xa
c0
=
cn
lim
xa
xn
+
?
?
?
+
c1
lim
xa
x
+
c0
= cnan + ? ? ? + c1a + c0
= p(a)
(rule 4) (rule 3) (rule 5 (repeatedly) and 2)
If
r
is
rational,
then
r(x)
=
p(x) q(x)
for
some
polynomials
p, q.
Rule
6
now
finished
things
off.
1If you end up with an indeterminate later using l'Ho^ pital's Rule.
form
0 0
,
,
-
,
etc.,
then
the
rules
don't
apply.
We
will
deal
with
these
limits
Examples
1.
Suppose
that
lim
xa
f
(x)
=
3,
lim
xa
g(x)
=
-1,
lim
xa-
h(x)
=
,
and
lim
xa+
h(x)
=
6.
Then
lim
xa
f
(x)
+
3g(x)
=
3
+
3(-1)
=
0
lim 2 f (x)g(x)h(x) = 2 ? 3 ? (-1) lim h(x) = -
xa-
xa-
lim 2 f (x)g(x)h(x) = 2 ? 3 ? (-1) ? 6 = 36
xa+
lim
xa-
g(x) f (x)h(x)
=
3
-1 lim h(x)
=
0
xa-
lim
xa+
g(x) f (x)h(x)
=
-1 3?6
=
-
1 18
2.
Simple evaluation:
lim
x1
x3
+ 2x2 - x 4x2 - 1
-
1
=
1+2-1-1 4-1
=
1 3
3.
Factorizing:
lim
x2-
x2 - 7x + 10 x2 - 4x + 4
=
lim
x2-
(x (x
- -
2)(x 2)(x
- -
5) 2)
= lim
x2-
x-5 x-2
= -3 lim
x2-
1 x-2
=
Roots and Rationalizing
Theorem.
lim
xa
f (x)
=
L
=
lim
xa
n
f (x) = n L
Recall how you would convert an expression with surds in the denominator into one with surds
in the numerator:
3
+45
=
3
+45
?
3 3
- -
55
=
4(3 - 5) 9-5
=
3
-
5
A similar approach can be used for limits.
Examples
1.
lim
x0
x
+
3 x
-
3
yields the indeterminate form
0 0
.
Multiplying by 1 =
xx
+ +
3 3
+ +
33
= 1 fixes
the problem:
lim
x0
x
+3 x
-
3
=
lim
x0
x
+3- x
3
?
xx
+ +
3 3
+ +
33
=
lim
x0
x(xx++33-+33)
=
lim
x0
x
1 +3+
3
=
21 3
2.
lim
x4
x2 + 9 - x-4
5
=
4 5
Comparing Limits and the Squeeze Theorem
While simple limits can be computed using the basic limit laws, more complicated functions are often best treated by comparison.
Theorem. Suppose that f (x) g(x) for all x = a and suppose that xlima f (x) and xlima g(x) both exist. Then
lim
xa
f
(x)
lim
xa
g(x)
Theorem
(Squeeze
Theorem).
Suppose
that
f (x)
g(x)
h(x)
for
all
x
=
a,
and
that
lim
xa
f
(x)
=
xlima h(x) = L. Then xlima g(x) exists and also equals L.
y
h(x)
g(x)
f (x)
a
x
y
Example In this example we compare the complicated
1
function f (x) = | that
xg|(. xS)in=cex-s1inx12sinwix1t2h
the much simpler 1 for all x = 0,
function we have
- |x| x sin
1 x2
|x|
Since lim |x| = 0 it follows that
x0
-1
1
x
lim x sin
x0
1 x2
=0
-1
Homework
1.
(a) (b)
PHreonvceetohratotxh-erwy =ise(cxo1/m3p-utye1/t3h)e(xli2m/3it+xlimx18/33xyx1-/-38+2 y2/3).
2. Suppose that f (x) < g(x) for all x = a and that limits of f and g both exist at x = a. Give
an
example
which
shows
that
we
may
only
conclude
that
lim
xa
f (x)
lim
xa
g(x).
That is, the
inequality need not be strict.
3.
Show
that
lim
x0
x2
cos(
1 x
)
exists
and
compute
it.
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