Calculating Limits
Theorem.
Suppose lim
x
→
a
f
(
x
)=
L
and lim
x
→
a
g
(
x
)=
M
.Le
t
c
be a constant. Then
1. lim
x
→
a
±
f
(
x
)+
g
(
x
)
²
=
L
+
M
2. lim
x
→
a
±
f
(
x
)
−
g
(
x
)
²
=
L
−
M
3. lim
x
→
a
±
cf
(
x
)
²
=
cL
4. lim
x
→
a
±
f
(
x
)
g
(
x
)
²
=
LM
5. lim
x
→
a
f
(
x
)
g
(
x
)
=
L
M
provided
M
6
=0.
For example, if lim
x
→
5
f
(
x
) = 7 and lim
x
→
5
g
(
x
) = 8 then lim
x
→
5
f
(
x
)
g
(
x
)
=
7
8
. W
ecanu
sethe
se
theorems to compute many limits.
Example.
It is not hard to believe that lim
x
→
4
x
= 4 or more generally that lim
x
→
a
x
=
a
.(The
closer
x
gets to
a
, the closer
x
gets to
a
.) So by the theorem, lim
x
→
4
x
2
= 16, lim
x
→
4
x
3
= 64,
etc.
More generally, lim
x
→
a
x
n
=
a
n
. So by the theorem, if
c
is any constant, lim
x
→
a
cx
n
=
ca
n
. E.g.,
lim
x
→
4
5
x
3
=5(4
3
) = 320
.
So we can handle constant multiples of powers of
x
. This means by the theorem we can take
limits of
polynomials
. Example: lim
x
→
4
(
5
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 Spring '08
 JohnRayko
 Calculus, Limits, lim, lim P, 5 g, x→a

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