The limit of f (x), as x approaches a, equals L
written: lim f ( x)  L
xa
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to a.
y
y  f ( x)
L
a
x
3x if x  2
Ex. lim f ( x) where f ( x)  
x 2
1 if x  2
y
lim f ( x) = lim  3x
x2
6
 3 lim x
x 2
 3(2)  6
Note: f (-2) = 1
is not involved
x2
x
-2
Suppose lim f ( x)  L and lim g ( x)  M
xa
xa
Then,
1. lim  f ( x)   Lr
r
xa
r , a real number
2. lim cf ( x)  c lim f ( x)  cL c, a real number
xa
xa
3. lim  f ( x)  g ( x)   L  M
x a
4. lim  f ( x) g ( x)   LM
xa
f ( x) L
f ( x) lim
5. lim
 xa

x a g ( x)
lim g ( x) M
xa
Provided that M  0
x 2  lim1
Ex. lim  x2  1  lim
x 3
x 3
x3
   lim1
 lim x
2
x 3
x 3
 32  1  10
lim  2 x  1
2 lim x  lim1
2x 1
x 1
x 1
x 1
Ex. lim


x 1 3 x  5
lim  3 x  5 
3lim x  lim 5
x 1
x 1
x 1
2 1 1


35 8
.
0
0
x5
Ex. xlim
5 x 2  25
0
Notice form
0
x5
 lim
x5  x  5  x  5 
1
1
 lim

x 5  x  5 
10
Factor and cancel
common factors
1
1
For all n > 0, lim n  lim n  0
x x
x x
1
provided that n is defined.
x
5 1
 2
2
3x  5 x  1
Divide
x
x
 lim
Ex. xlim
2
2
2
x


by x
2  4x

4
x2
5
 1 
lim 3  lim    lim  2 
x 
x   x  x   x 
3 0 0
3



 2 
04
4
lim  2   lim 4
x   x  x 
.
3
The right-hand limit of f (x), as x approaches a,
equals L
written: lim f ( x)  L
x a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
right of a.
y  f ( x)
L
a
The left-hand limit of f (x), as x approaches a,
equals M
written: lim f ( x)  M
x a
if we can make the value f (x) arbitrarily close
to M by taking x to be sufficiently close to the
y
left of a.
y  f ( x)
M
a
.
x
 x 2 if x  3
Ex. Given f ( x)  
2x if x  3
Find lim f ( x)
x3
lim f ( x)  lim 2 x  6
x 3
x 3
Find lim f ( x)
x3
lim f ( x)  lim x2  9
x3
x3