The document discusses limits and continuity. It defines a limit as describing the behavior of a function as the input value approaches a particular number. It provides examples of calculating limits using direct substitution, factorizing, and special limits involving infinity. The key points covered are: - A limit describes what value a function approaches as its input gets closer to a number - Methods for calculating limits include direct substitution, factorizing, and using special rules for infinity - A function is continuous if the left-hand and right-hand limits are equal at a point

1. Limits & Continuity

2. Limits & Continuity
A limit describes the behaviour of functions.

3. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  :
x a

4. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a

5. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  :
x a

6. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?

7. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
y
y  x 1
1
1 x

8. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
y
y  x 1
1
1 x
lim x  1  0

x1

9. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

x1

10. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim f  x   4

x1 x0

11. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
lim f  x  :
x a
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim f  x   4

x1 x0

12. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
lim f  x  : as the x value approaches a from the positive side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim f  x   4

x1 x0

13. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
lim f  x  : as the x value approaches a from the positive side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim x  1  0

lim f  x   4

x1 x1 x0

14. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
lim f  x  : as the x value approaches a from the positive side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim x  1  0

lim f  x   4

lim f  x   6

x1 x1 x0 x0

15. Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  : as the x value approaches a, what value does f(x) approach?
x a
lim f  x  : as the x value approaches a from the negative side,
x a
what value does f(x) approach?
lim f  x  : as the x value approaches a from the positive side,
x a
what value does f(x) approach?
y y
y  x 1 6
4 y  f  x
1
1 x x
lim x  1  0

lim x  1  0

lim f  x   4

lim f  x   6

x1 x1 x0 x0
If lim f  x   lim f  x  , then f  x  is continuous at x  a
x a x a

16. Finding Limits

17. Finding Limits
(1) Direct Substitution

18. Finding Limits
(1) Direct Substitution
e.g. lim x  7
x5

19. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12

20. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12
(2) Factorise and Cancel

21. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12
(2) Factorise and Cancel
x2  9
e.g. lim
x3 x  3

22. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12
(2) Factorise and Cancel
e.g. lim
x2  9
 lim
 x  3 x  3
x3 x  3 x3  x  3

23. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12
(2) Factorise and Cancel
e.g. lim
x2  9
 lim
 x  3 x  3
x3 x  3 x3  x  3
 lim  x  3
x3

24. Finding Limits
(1) Direct Substitution
e.g. lim x  7  5  7
x5
 12
(2) Factorise and Cancel
e.g. lim
x2  9
 lim
 x  3 x  3
x3 x  3 x3  x  3
 lim  x  3
x3
 33
6

25. (3) Special Limit

26. (3) Special Limit
1
lim  0
x x

27. (3) Special Limit
1
lim  0
x x
x3  3x 2  2 x  1
e.g.  i  lim
x 4 x3  1

28. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x

29. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x

30. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x
1

4

31. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x
1

4
4x  x2
 ii  lim 3
x x  1

32. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x
1

4
4x  x2 0
 ii  lim 3
x x  1

1
0

33. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x
1

4
4x  x2 0 x7  x6  x2
 ii  lim 3
x x  1
  iii  lim 7
x 3 x  x  974
1
0

34. (3) Special Limit
1
lim  0
x x
x3 3x 2 2 x 1
x3  3x 2  2 x  1  3  3 3
e.g.  i 
3
lim  lim x x x x
x 4x 1
3
x 4 x3 1
3
 3
x x
3 2 1
1  2  3
 lim x x x
x 1
4 3
x
1

4
4x  x2 0 x7  x6  x2 1
 ii  lim 3
x x  1
  iii  lim 7
x 3 x  x  974

1 3
0

35. x3  2
 iv  lim 2
x x  1

36. x3  2 1
 iv  lim 2
x x  1

0


37. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1

38. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1
lim
 x  3 x  2 
x  x  1 x  1

39. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1
 x  3 x  2  x2  x  6
lim  lim 2
x  x  1 x  1 x x 1

40. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1
 x  3 x  2  x2  x  6
lim  lim 2
x  x  1 x  1 x x 1
1

1
1

41. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1
 x  3 x  2  x2  x  6
lim  lim 2
x  x  1 x  1 x x 1
1

1
1
 horizontal asymptote is y  1

42. x3  2 1
 iv  lim 2
x x  1

0

 x  3 x  2 
 v  Find the horizontal asymptote of y 
 x  1 x  1
 x  3 x  2  x2  x  6
lim  lim 2
x  x  1 x  1 x x 1
1

1
1
 horizontal asymptote is y  1 Exercise 7I; 1a, 2ace, 3ac,
4a, 5ad, 8a, 9ab, 10a