Limits of functions ( part I ) and continuity

Definitions and properties
Special techniques to calculate lim f(x)
Continuity

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 {x_n} such that

 
        lim x_n = b ;
        for each n, x_n is different from b ;
        for each n, x_n is in the domain of f.

With each sequence {x_n} corresponds an 'image sequence' {f(x_n)}.
If, for all these image sequences, lim f(x_n) = (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

 
If      lim f(x)  not = lim f(x)    then  lim  f(x) is not defined
        < b             > b                b

 
If      lim f(x)  = lim f(x) = c    then  lim  f(x) = c
        < b         > b                    b

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.x^m + ... + k   is a polynomial then


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

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.

 
        f(x) is continuous in b
                <=>
        lim f(x) = f(b)
         b

 
        f(x) is left continuous in b
                <=>
        lim f(x) = f(b)
        < b

 
        f(x) is right continuous in b
                <=>
        lim f(x) = f(b)
        > b

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.

Take m = (a + b)/2 . If f(m) = 0 , the theorem is proved.
In the other case, there is an interval [a₁,b₁], with a₁ = m or b₁ = m, such that f(a₁) < 0 and f(b₁) > 0.
Again, take m₁ = (a₁ + b₁)/2 . If f(m₁) = 0 , the theorem is proved. .... After n steps we have [a_n,b_n] with f(a_n) < 0 and f(b_n) > 0.
The length of [a_n,b_n] is
```
 
        (b-a)
        ------
          2ⁿ
```
In that way we have the sequences
a =< a₁ =< a₂ =< a₃ =< a₄ ...
b >= b₁ >= b₂ >= b₃ >= b₄ ...
The first sequence is not descending and has an upper bound. So lim a_n = c₁.
The second sequence is not rising and has a lower bound. So lim b_n = c₂.
But,
```
 
                    (b-a)
lim (b_n-a_n) = lim  ----- = 0     => lim a_n = lim b_n =c₁ = c₂ = c
infty        infty   2ⁿ
```
```
 
Now, Since f(x) is continuous in c , lim f(x) = f(c)
                                      c
```
Hence, for each sequence {x_n} with limit c, the sequence {f(x_n)} has limit f(c). Thus,
```
 
        lim  f(a_n)  = f(c)  = lim f(b_n)
       infty                 infty
```
All terms f(a_n) are < 0 , thus f(c) =< 0 .
All terms f(b_n) are > 0 , thus f(c) >= 0 .
So, f(c) must be 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

Other tutorials about the same subject

Limits of functions (pdf)

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