Limits & Continuity
Limits & ContinuityA limit describes the behaviour of functions.
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
( )lim :x a
f x+→
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
( )lim :x a
f x+→
as the x value approaches a from the positive side, what value does f(x) approach?
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
( )lim :x a
f x+→
as the x value approaches a from the positive side, what value does f(x) approach?
1lim 1 0x
x+→
− =
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
( )lim :x a
f x+→
as the x value approaches a from the positive side, what value does f(x) approach?
1lim 1 0x
x+→
− = ( )0
lim 6x
f x+→
=
Limits & ContinuityA limit describes the behaviour of functions.
( )lim :x af x
→as the x value approaches a, what value does f(x) approach?
( )lim :x a
f x−→
as the x value approaches a from the negative side, what value does f(x) approach?
y
x1
1
1y x= −
1lim 1 0x
x−→
− =
y
x
46
( )y f x=
( )0
lim 4x
f x−→
=
( )lim :x a
f x+→
as the x value approaches a from the positive side, what value does f(x) approach?
1lim 1 0x
x+→
− = ( )0
lim 6x
f x+→
=
( ) ( ) ( )If lim lim , then is continuous at x a x a
f x f x f x x a− +→ →
= =
Finding Limits
Finding Limits(1) Direct Substitution
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
(2) Factorise and Cancel
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
(2) Factorise and Cancel2
3
9e.g. lim
3x
xx→
−−
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
(2) Factorise and Cancel2
3
9e.g. lim
3x
xx→
−−
( ) ( )( )3
3 3lim
3x
x x
x→
+ −=
−
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
(2) Factorise and Cancel2
3
9e.g. lim
3x
xx→
−−
( ) ( )( )3
3 3lim
3x
x x
x→
+ −=
−( )
3lim 3x
x→
= +
Finding Limits(1) Direct Substitution
5e.g. lim 7
xx
→+ 5 7
12
= +=
(2) Factorise and Cancel2
3
9e.g. lim
3x
xx→
−−
( ) ( )( )3
3 3lim
3x
x x
x→
+ −=
−( )
3lim 3x
x→
= +
3 3
6
= +=
(3) Special Limit
(3) Special Limit1
lim 0x x→∞
=
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−14
=
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−14
=
( )2
3
4 lim
1x
x xii
x→∞
−+
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−14
=
( )2
3
4 lim
1x
x xii
x→∞
−+
010
=
=
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−14
=
( )2
3
4 lim
1x
x xii
x→∞
−+
010
=
=
( )7 6 2
7 lim3 974x
x x xiii
x x→∞
+ +− −
(3) Special Limit1
lim 0x x→∞
=
( )3 2
3
3 2 1e.g. lim
4 1x
x x xi
x→∞
+ + −−
3 2
3 3 3 3
3
3 3
3 2 1
lim4 1x
x x xx x x x
xx x
→∞
+ + −=
−
2 3
3
3 2 11
lim1
4x
x x x
x
→∞
+ + −=
−14
=
( )2
3
4 lim
1x
x xii
x→∞
−+
010
=
=
( )7 6 2
7 lim3 974x
x x xiii
x x→∞
+ +− −
13
=
( )3
2
2 lim
1x
xiv
x→∞
+−
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +( ) ( )( ) ( )
3 2lim
1 1x
x x
x x→∞
+ −− +
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +( ) ( )( ) ( )
3 2lim
1 1x
x x
x x→∞
+ −− +
2
2
6lim
1x
x xx→∞
+ −=−
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +( ) ( )( ) ( )
3 2lim
1 1x
x x
x x→∞
+ −− +
2
2
6lim
1x
x xx→∞
+ −=−
111
=
=
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +( ) ( )( ) ( )
3 2lim
1 1x
x x
x x→∞
+ −− +
2
2
6lim
1x
x xx→∞
+ −=−
111
=
=horizontal asymptote is 1y∴ =
( )3
2
2 lim
1x
xiv
x→∞
+−
10
=
= ∞
( ) ( ) ( )( ) ( )
3 2 Find the horizontal asymptote of
1 1
x xv y
x x
+ −=
− +( ) ( )( ) ( )
3 2lim
1 1x
x x
x x→∞
+ −− +
2
2
6lim
1x
x xx→∞
+ −=−
111
=
=horizontal asymptote is 1y∴ = Exercise 7I; 1a, 2ace, 3ac,
4a, 5ad, 8a, 9ab, 10a