Limits of functions ( part I ) and continuity




Definitions and properties

Definition

Say f: R -> R : x -> f(x) is a real function and a value b.
If b is a real number, assume that f(x) is defined in ]b-e,b+e[ or ]b-e,b[ or ]b,b+e[ .
If b is +infinity, assume that f(x) is defined for all x > a fix number N.
If b is -infinity, assume that f(x) is defined for all x < a fix number N.

Take the set of all sequences {xn} such that

 
        lim xn = b ;
        for each n, xn is different from b ;
        for each n, xn is in the domain of f.
With each sequence {xn} corresponds an 'image sequence' {f(xn)}.
If, for all these image sequences, lim f(xn) = (a fixed value c), then we say that lim f(x) = c . We write
 
        lim f(x) = c   or    lim f(x) = c
       x->b                   b
If there isn't such value c, we say that lim f(x) is not defined.

Examples

 
             x.x - 5x + 6           (x - 2)(x - 3)
        lim --------------- =  lim --------------- = 1
         3      x - 3           3      x - 3


        lim sqrt(x) is not defined
        -2

               2.x2 + x       2
        lim  ------------  = ---
      +infty   3.x2 + 4       3

Properties

If both sides exist, it can be proved that :
 
        lim c.f(x) = c . lim f(x)      (with c = constant)
         b                b

        lim |f(x)|= |lim f(x)|
         b            b

                p              p
        lim f(x)  = (lim f(x) )
         b            b

        lim pth-root(f(x)) = pth-root(lim f(x) )
         b                             b


        lim (f(x) + g(x)) = lim f(x)  + lim g(x)
         b                   b           b

        lim (f(x) . g(x)) = lim f(x)  . lim g(x)
         b                   b           b

        lim (f(x) / g(x)) = lim f(x)  / lim g(x)
         b                   b           b

Some special limits

If both sides exist, we have
 
        lim r = r  (for each constant number r)
         b

        lim x = b
         b
             n     n
        lim x  = b       (n is positive integer)
         b

        lim (1/x) = 0
       infty
                 n
        lim (1/x) = 0    (n is positive integer)
       infty


        lim nth-root(x) = +infty
      +infty

        lim nth-root(x) = -infty
      -infty


        lim 1/nth-root(x) = 0
      +infty

      If T(x) and N(x) are polynomials

        lim T(x) = T(b)
         b

            T(x)     T(b)
        lim ---- =   ----
         b  N(x)     N(b)


        lim sqrt(T(x)) = sqrt(T(b))
         b

Left and right limit

Say f: R -> R : x -> f(x) is a real function and there is a strictly positive real number e such that ]b-e,b[ is part of the domain of the function f.
We restrict the domain of f to ]b-e,b[.
With this new domain, we take
 
        lim f(x)
         b
This limit is called 'the left limit of f(x) in b' .
This left limit is noted
 
        lim f(x)
        < b
Analogous: f: R -> R : x -> f(x) is a real function and there is a strictly positive real number e such that ]b,b+e[ is part of the domain of the function f.
We restrict the domain of f to ]b,b+e[.
With this new domain, we take
 
        lim f(x)
         b
This limit is called 'the right limit of f(x) in b' .
This right limit is noted
 
        lim f(x)
        > b

Important example

 
              1
        lim ------ = - infty
        < b  x - b

              1
        lim ------ = + infty
        > b  x - b

Corollary

Remark :
 

If  lim  f(x) = c  then we have not always  lim f(x)  = lim f(x) = c
     b                                      < b         > b

Special techniques to calculate lim f(x)

The case +infty - infty with polynomials

Example

 
         lim (2x2  - 6x + 7) =
        infty

         lim  2x2 .(1 - 3/(2x) + 7/(2x.x) ) =
        infty

         lim  2x2 .  lim (1 - 3/(2x) + 7/(2x.x) ) =
        infty      infty

         lim  2x2
        infty

Rule:

 
If T(x) = a.xn + ... + l   is a polynomial then

         lim  T(x) = lim  (a.xn )
        infty        infty

The case infty / infty with a quotient of polynomials

In the same way as above you can prove that
 
If T(x) = a.xn + ... + l   is a polynomial then

And  N(x) = b.xm + ... + k   is a polynomial then


              T(x)          ( a.xn )
         lim  ----- = lim   -------
        infty N(x)    infty ( b.xm )

k/0 with a quotient of polynomials

Example:
 
             -5x - 81              -5x - 81       1
        lim  -------------- = lim  --------- .  -------
        > -3 (x + 3)(x - 1)  > -3   (x - 1)     (x + 3)



        = (16.5) . (+infty) = (+infty)

----------

             -5x - 81              -5x - 81       1
        lim  -------------- = lim  --------- .  -------
        < -3 (x + 3)(x - 1)  < -3   (x - 1)     (x + 3)



        = (16.5) . (-infty) = (-infty)

0/0 with a quotient of polynomials

Example 1:
 

             2x.x - 4x              2x.(x - 2)
        lim -------------- =  lim ---------------
        > 2  x.x - 4x + 4     >  2  (x - 2)(x - 2)


                  2x
        = lim --------- = +infty
          > 2  (x - 2)

Example 2:
 

             2x.x - 4x              2x.(x - 2)
        lim -------------- =  lim ---------------
         2   x.x - 5x + 6      2   (x - 2)(x - 3)


                  2x
        = lim --------- = -4
           2   (x - 3)

0/0 with irrational functions

Example 1:
 
            sqrt(x-3) -1
        lim ------------- =
         4     x - 4


            (sqrt(x-3) -1)(sqrt(x-3) +1)
        lim ---------------------------- =
         4    (x - 4)    (sqrt(x-3) +1)


                 (x - 3 - 1)
        lim ---------------------------- =
         4    (x - 4)    (sqrt(x-3) +1)

                 1
        lim ---------------- = 0.5
         4    (sqrt(x-3) +1)
Example 2:
 
                  ______________
                 |  2
            1 - \| x  - 3 x + 3
        lim ---------------------
         1      _________
               |    2
              \| 4 x  - 3  - 1

We multiply the numerator and the denominator with the factor  F =
                _____________      _________
               |  2               |    2
        ( 1 + \| x  - 3 x + 3)(  \| 4 x  - 3  + 1 )


                                     _________
                  2                 |    2
            (1 - x  + 3 x - 3)  (  \| 4 x  - 3  + 1 )
      = lim --------------------------------------------------------
         1                          _____________
                  2                |  2
             ( 4 x  - 4 )   ( 1 + \| x  - 3 x + 3)



            (1 - x2  + 3 x - 3)
      = lim ---------------------- = ... = 1/8
         1    ( 4 x2  - 4 )
Example 3
 
                 ______________     _________
               3|  3              3|    2
               \| x  - 2 x - 3  - \| 2 x  - 7
          lim ----------------------------------
           2      2 x3  + x - 18


We multiply the numerator and the denominator with the factor  F =

  _________________     ___________________________     _____________
3|   3           2    3|   3                2         3|     2     2
\| (x  - 2 x - 3)   + \| (x  - 2 x - 3) (2 x  - 7)  + \| (2 x  - 7)


Then we have for the limit

              x3 - 2 x - 3 - 2 x2 + 7
        lim ----------------------------
          2    (2 x3 + x - 18) . F

             (x2 - 2) (x - 2)
       =lim ----------------------------
          2  (2 x2 + 4 x + 9) (x - 2) . F

             (x2 - 2)                    2
       =lim ----------------------- = --------
          2  (2 x2 + 4 x + 9)  . F      25 .3

k/0 with irrational functions

Example 1:
 
             1 + sqrt(-x)
        lim -------------- =
       > -2     x + 2

            (1 + sqrt(-x))     1
        lim ---------------.-------- = +infty
       >-2      1            (x + 2)
Example 2:
 
             x2 - 5x + 4
        lim ------------------- =
        >3   sqrt(x2 - 5x + 6)

                                   1
        lim (x2 - 5x + 4).------------------- =
        >3                 sqrt( (x-2)(x-3) )


        (-2).(+infty) = -infty
Example 3:
 
             x2  - 5 x + 4
        lim ----------------- =
        <3    _____________
             |  2
            \| x  - 5 x + 6

                                   1
        lim (x2 - 5x + 4).------------------- =
        <3                 sqrt( (x-2)(x-3) )


        is not defined

infty/infty with irrational functions

Example 1:
 
                  ________
                 |  2
                \| x  + 1  + 3 x
        lim     ------------------- =
        +infty      2x - 5


                   _________
                  |      -2
                (\| 1 + x    + 3) x
        lim     ----------------------- =
        +infty     x.( 2 - 5/x)


                     _________
                    |      -2
                  (\| 1 + x    + 3)           4
        lim     -------------------------- = --- = 2
        +infty     ( 2 - 5/x)                 2
Example 2:
 
                   ________
                  |  2
                 \| x  + 1  + 3 x
        lim     ------------------- =
        -infty      2x - 5



                          _________
                         |      -2
                  x(3 - \| 1 + x   )
        lim     ------------------------ =
        -infty     x.( 2 - 5/x)


                           _________
                          |      -2
                    (3 - \| 1 + x   )         2
        lim     -------------------------- = --- = 1
        -infty     ( 2 - 5/x)                 2

+infty - infty with irrational functions

Example 1:
 
                  ________________
                 |    2
        lim (   \| 4 x  + 3 x - 1  + 2 x  ) =
       -infty

               ________________          ________________
              |    2                    |    2
            (\| 4 x  + 3 x - 1  + 2 x)(\| 4 x  + 3 x - 1  - 2 x)
        lim -----------------------------------------------------  =
       -infty           ________________
                       |    2
                     (\| 4 x  + 3 x - 1  - 2 x)



             (4x2 + 3x - 1) - 4x2
        lim -------------------------------------  =
       -infty        ________________
                    |    2
                  (\| 4 x  + 3 x - 1  - 2 x)


                   ( 3x - 1)
        lim --------------------------------  =
       -infty     ________________
                 |    2
               (\| 4 x  + 3 x - 1  - 2 x)


                  x ( 3 - 1/x)
        lim --------------------------------  =
       -infty       _____________
                   |     3    -2
               (-  | 4 + - - x    - 2) x
                  \|     x


                ( 3 - 1/x)                       3
        lim --------------------------------  = ----
       -infty       _____________                4
                   |     3    -2
               (-  | 4 + - - x    - 2)
                  \|     x

Example 2:
 
                    _____________
                   |    2
        lim ( 5 + \| 4 x  - x + 3 + 2 x )
       -infty
                      _____________
                     |    2
     =  lim (2 x +  \| 4 x  - x + 3  )   + 5
       -infty

                      _____________             _____________
                     |    2                    |    2
            (2 x +  \| 4 x  - x + 3  )(2 x -  \| 4 x  - x + 3 )
     =  lim ---------------------------------------------------  + 5
       -infty                  _____________
                              |    2
                     (2 x -  \| 4 x  - x + 3  )

              x - 3
     =  lim ------------------------------ +  5
       -infty           _____________
                       |    2
              (2 x -  \| 4 x  - x + 3  )

             x( 1 - 3/x )
     =  lim ------------------------------ +  5
       -infty           _________________
                       |              2
             x (2  +  \| 4 - 1/x + 3/x    )

             ( 1 - 3/x )
     =  lim ------------------------------ +  5  = 1/4 + 5
       -infty           _________________
                       |              2
               (2  +  \| 4 - 1/x + 3/x    )

Many other special limits

For calculation of many other special limits (and examples) using derivatives see
Two special limits and L'Hospitals rule and Examples .

Continuity

Definitions

Let f(x) be a real function.

Corollaries

If f(x) is left continuous and right continuous in b, then f(x) is continuous in b

Is f+g continuous?

If f and g are continuous in b, then f+g is continuous in b. Proof:
 
f is continuous in b => lim f(x) = f(b)
                         b

g is continuous in b => lim g(x) = g(b)
                         b
So,

lim (f(x) + g(x)) = lim f(x) + lim g(x) = f(b) + g(b)
 b                   b          b
Q.E.D. ..
In the same way it you can prove that : If f and g are continuous in b, then f-g is continuous in b. If f and g are continuous in b, then f.g is continuous in b. If f and g are continuous in b, then f/g is continuous in b. (with g(b) not 0)

Theorem

It can be proved that all algebraic and trigonometric functions are continuous in all elements of their domain.

Continuity of f(x) in an interval.

 
        f(x) is continuous in an interval [a,b]
                <=>
        f(x) is continuous in each element of [a,b]

Bolzano's theorem

If f(x) is continuous in [a,b] and f(a).f(b) < 0
Then there is a real number c in ]a,b[ such that f(c) = 0.
Proof:
We'll prove the theorem for f(a) < 0 and f(b) > 0.
For another proof see Theoretical part

Theorem of the mean values.

If f(x) is continuous in [a,b] and r is a real number between f(a) and f(b), then there is a number c in ]a,b[ such that f(c) = r.
Proof:
Construct the function g(x) = f(x) - r.
Since f(x) and r are both continuous in [a,b], g(x) is continuous in [a,b].
Since r is a real number between f(a) and f(b), 0 is a real number between g(a) and g(b).
So, we can apply the Bolzano theorem on g(x) in [a,b].
Hence, there is a number c in ]a,b[ such that g(c) = 0.
This means that there is a number c in ]a,b[ such that f(c) = r.

Theorem of Weierstrass.

If f is continuous in [a,b] , then f(x) attains a maximal and a minimal image in [a,b]. For a proof of the theorem see Theoretical part




Topics and Problems

MATH-abundance home page - tutorial

MATH-tutorial Index

The tutorial address is http://home.scarlet.be/math/

Copying Conditions

Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be     The subject of the mail must contain the flemish word 'wiskunde' because other mails are filtered to Trash