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Limits & Continuity
Limits & Continuity
A limit describes the behaviour of functions.
Limits & Continuity
A limit describes the behaviour of functions.
lim f  x  :
x a
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
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
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?
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
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
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
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
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
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
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
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
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
Finding Limits
Finding Limits
(1) Direct Substitution
Finding Limits
(1) Direct Substitution
   e.g. lim x  7
        x5
Finding Limits
(1) Direct Substitution
   e.g. lim x  7  5  7
        x5
                   12
Finding Limits
(1) Direct Substitution
   e.g. lim x  7  5  7
        x5
                   12

(2) Factorise and Cancel
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
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
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
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
(3) Special Limit
(3) Special Limit
                        1
                    lim  0
                    x x
(3) Special Limit
                                 1
                             lim  0
                             x x


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

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11X1 T09 01 limits & continuity (2011)

  • 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
  • 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
  • 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