Asymptotes




An asymptote

A line d is an asymptote of a curve C if and only if the limit of the distance from a point P of C to d is zero, if P recedes indefinitely along the curve.
We say that C has an asymptote d.

Asymptotes and functions.

If the graph of a function has an asymptote d, then we say that the function has an asymptote d.
A function can have more than one asymptote.
If an asymptote is parallel with the y-axis, we call it a vertical asymptote. If an asymptote is parallel with the x-axis, we call it a horizontal asymptote. All other asymptotes are oblique asymptotes.
In this article we only consider functions that are continuous in their domain.

Vertical asymptotes

From the definition we have :
 
A line x = a is a vertical asymptote of a function f(x)
                <=>
(lim f(x) = +infty  or -infty ) or (lim f(x) = +infty  or -infty )
 > a                                < a

Examples

 
     x+4                                                   x+4
lim ----- = -infty , so x = 3  is a vertical asymptote of -----
< 3  x-3                                                   x-3

     x+4                                                   x+4
lim ----- = -infty , so x = 3  is a vertical asymptote of -----
> 3  x-3                                                   x-3

     x.x + 3x + 2
lim -------------- = -1 , so x = -2 is not a vertical asymptote.
-2      x + 2


 lim    tan(x) = +infty , so x = pi/2  is a vertical asymptote of tan(x)
< pi/2

The function tan(x) has many vertical asymptotes.

Horizontal asymptotes -- general formulas

From the definition we have :

 
A line y = b is a horizontal asymptote of a function f(x)
                <=>
 lim f(x) = b  or   lim f(x) = b    with b in R
+infty             -infty

Examples :

 
     3x2 - 4x -1
lim  -------------- = 0.5
infty   6x2 - 6
                                                    3x2 - 4x -1
So, y = 0.5 is a horizontal asymptote of a function -------------
                                                       6x2 - 6

          _______                         _______
         |  2                            |  2
        \| x  - 1                       \| x  - 1
 lim   -------------- = 1  and   lim   -------------- = -1
+infty    x - 1                 -infty    x - 1

So, y = 1 and y = -1 are horizontal asymptotes.

Oblique asymptotes

General formulas

Each oblique asymptote d has an equation y = ax + b. Here a and b are unknown real numbers. We'll deduce formulas to calculate a and b from the function f(x).

 
        y =  ax + b is oblique asymptote d

                <=>

                lim |P,Q| = 0
             P->infty

                <=>

                lim |P,Q| = 0
              x-> infty
                              in triangle PQS is |P,Q|=|P,S|.sin(PSQ)
                <=>

                lim |P,S|.sin(PSQ) = 0
              x-> infty

                        since sin(PSQ) is constant and not zero
                <=>

                lim |P,S| = 0
              x-> infty

                        since |P,S| = |f(x) - ax - b|
                <=>

                 lim  |f(x) - ax - b| = 0
              x-> infty

                <=>

                 lim  f(x) - ax - b = 0         (*)
              x-> infty

From (*) we'll deduce a formula for a
 
                    f(x) - ax - b
        (*) => lim ----------------- = 0
              infty      x


                        <=>

                    f(x)                b
                lim ---- - lim a - lim ---  = 0
             infty   x                  x

                        <=>


                     f(x)
                (lim ---- ) -  a = 0
               infty  x

                        <=>


                    f(x)
                lim ----  =  a                  (1)
              infty  x

With (*) we'll construct a formula for b.
 
        (*) => lim (f(x) - ax ) = b             (2)
               infty
Since we know a from (1), we can calculate b from (2).

 
    y = ax + b is an oblique asymptote of the curve  y=f(x)

                 if and only if

                    f(x)
            a = lim ----
              infty  x

    and     b =  lim (f(x) - ax )
                 infty

About oblique asymptote of a rational function

We represent the rational function f (x) by t (x) / n (x). t(x) and n(x) are polynomials in x.
  1. If the degree of the numerator t(x) is at least two units higher than the degree of the denominator n(x) then we have
     
                        f(x)
                    lim ---- = O
                  infty  x
    
    There is no oblique asymptote for f(x).
  2. If the degree of the numerator t(x) is less than or equal to the degree of the denominator n(x) then we have
     
                        f(x)
                    lim ---- = O
                  infty  x
    
    There is no oblique asymptote for f(x).
  3. If the degree of the numerator t(x) is just one unit higher than the degree of the denominator n(x) then we have
     
                        f(x)
               a =  lim ---- = a finite number , not 0
                  infty  x
    
    There is an oblique asymptote for f(x).

Concrete examples:

 
   3x3 + 2x +1
   -------------  has no oblique asymptote
     6 x + 5

   3x3 + 2x +1
   ---------------  has no oblique asymptote
     6 x3 + 5x


   3x3 + 2x +1
   -------------  has an oblique asymptote
     6 x2 + 5

   We calculate oblique asymptote

             3x3 + 2x +1
   a = lim ------------------ = 1/2
      infty   6 x3 + 5x


               3 x3 + 2x +1          x
   b =  lim ( ----------------- -  ----- )
       infty   6 x2 + 5              2



                -x  + 2
     =  lim  -------------- = 0
       infty        2
                12 x  + 10

   The oblique asymptote  is y = x/2

Simple method for a rational function.

We assume here that the oblique asymptote exists. So, the degree of the numerator is just one unit higher than the degree of the denominator.

One can calculate the oblique asymptote of the rational function t (x) / n (x) in a different way as above.

To this end one makes the Euclidean division of t(x) by n(x). We will now show that the quotient of that division constitutes the oblique asymptote.

Proof:

For dividend t (x), divisor n (x) quotient q (x) and remainder r (x) we have

 
           t(x)           r(x)
          ----- = q(x) + ------
           n(x)           n(x)
The quotient has a degree = 1 in x.

The graph of the rational function is the sum of two parts.
The first part is the line y = q(x). In the second part is the degree of r(x) smaller than the degree of n(x). This means that the limit of this second part is zero if x -> infty.
So, the graph of the given rational function approaches the line y= q(x) if x -> infty. This means that y = q(x) is the oblique asymptote.

Example: We retake

 
        3x3 + 2x +1
  y =   -------------
          6 x + 5

The quotient of the division of  x3 + 2x +1 by 6 x + 5 is x/2

The oblique asymptote is x/2
General rule:

A rational function has an oblique asymptote if and only if the degree of the numerator is just one unit higher than the degree of the denominator.
The Euclidean division of numerator by denominator gives a quotient ax+b.
The oblique asymptote of that rational function is y = ax+b.

Example 1 :

Find the oblique asymptote of
 
           7 x4 - 3 x2 + x
  f(x) = --------------------
          3 x3 - 2x2 + 4
The Euclidean division of numerator by denominator gives a quotient (7/3) x + 14/9.

The equation of the oblique asymptote is y = (7/3) x + 14/9.

Example 2

 
              ________
             |  2
Take f(x) = \| x  - 1  + 2
We calculate a and b.
First, let x -> + infty . Then ________ | 2 \| x - 1 + 2 a = lim ------------------ x __________ | \| 1 - 1/x2 + 2/x = lim ----------------------- = 1 1 ________ | 2 b = lim \| x - 1 + 2 - 1.x (sqrt(x2 -1) - x )( sqrt(x2 -1) + x) = lim --------------------------------------- + 2 sqrt(x2 -1) + x x2 - 1 - x2 = lim ----------------------------- + 2 sqrt(x2 -1) + x - 1 = lim --------------------- + 2 sqrt(x2 -1) + x = 2 So the asymptote is y = x + 2 if x -> + infty Next, let x -> - infty . Then ________ | 2 \| x - 1 + 2 a = lim ------------------ x __________ | - \| 1 - 1/x2 + 2/x = lim ----------------------- = -1 1 ________ | 2 b = lim \| x - 1 + 2 + 1.x (sqrt(x2 -1) + x )( sqrt(x2 -1) - x) = lim --------------------------------------- + 2 sqrt(x2 -1) - x x2 - 1 - x2 = lim ----------------------------- + 2 sqrt(x2 -1) - x -1 = lim --------------------- + 2 sqrt(x2 -1) + x = 2 So the asymptote is y = - x + 2 if x -> - infty
Alternative method:

If x tends to + infinity, so for very large values of x, the graphs of

 
              ________
             |  2
     f(x) = \| x  - 1  + 2
and
              _____
             |  2
     g(x) = \| x       + 2
are infinitely close to each other because the number -1 has almost no influence.
But g(x) is nothing else than the line y = x+2. So, y = x+2 is the asymptote for x -> + infty.

If x -> - infty , the graphs of

 
              ________
             |  2
     f(x) = \| x  - 1  + 2

and           _____
             |  2
     g(x) = \| x       + 2    with x negative.


          = - x + 2
are infinitely close to each other because the number -1 has almost no influence.
But g(x) is nothing else than the line y = - x+2. So, y = - x+2 is the asymptote for x -> - infty.

Example 3:

Find the oblique asymptote for x -> - infty of the function f(x) =

 
    x2 + x -2
   -----------------
   sqrt(x2 - x - 2)
The asymptote has the form y = ax+b. We calculate a and b.
All limits in this exercise are for x --> - infty.
 

             x2 + x -2
  a = lim -------------------
          x sqrt(x2 - x - 2)


           1 + 1/x - 2/x2
   =  lim ----------------------
          - sqrt(1 -1/x - 2/x2)

   = -1

              x2 + x -2
  b = lim (------------------- + x)
             sqrt(x2 - x - 2)

            x2 + x -2 + x sqrt(x2 - x - 2)
    = lim ---------------------------------
               sqrt(x2 - x - 2)


           ( x2 + x -2 + x sqrt(x2 - x - 2)) ( x2 + x -2 - x sqrt(x2 - x - 2))
    = lim --------------------------------------------------------------------------
           sqrt(x2 - x - 2) ( x2 + x -2 - x sqrt(x2 - x - 2))


           (x2 + x -2)2  - x2 (x2 - x - 2)
    = lim ----------------------------------------------------------
           sqrt(x2 - x - 2) ( x2 + x -2 - x sqrt(x2 - x - 2))


                 3 x3 - x2 - 4x -4
    = lim ----------------------------------------------------------
           sqrt(x2 - x - 2) ( x2 + x -2 - x sqrt(x2 - x - 2))

 We divide numerator and denominator by x3

             3 - 1/x - 4/x2 - 4/x3
    = lim --------------------------------------------------------------
           -sqrt(1 - 1/x - 2/x2) (1 + 1/x -2/x2 +sqrt( 1 - 1/x -2/x2))



    =  3/(-2)

The oblique asymptote is y = - x -3/2

Example 4:

Find the oblique asymptote of f(x) = (3x + 3x2 + x3)1/3 for x -> + infty.
Although this asymptote can be found by the ordinary method of computation, the calculations are fairly comprehensive. Therefore we show here two other methods. The first method uses a translation. The second method is a way which is less mathematically.

f(x) = (3x + 3x2 + x3)1/3
We know 1 + 3x + 3x2 + x3 = (1 + x)3 ; so,
f(x) = ( (x + 1)3 - 1)1/3
We shift the graph of f(x) one unit to the right. The new function is
g(x) = ( x3 - 1)1/3
We calculate the oblique asymptote y = ax + b of g(x) for x -> + infty.

 
         ( x3 - 1)1/3
 a = lim -------------
              x

            x3 - 1
   = lim (--------- )1/3
              x3

                 1
   = lim ( 1 - ---- )1/3
                x3

   = 1


 b = lim ( ( x3 - 1)1/3 - x )

We multiply numerator and denominator with

      ( x3 - 1)2/3 +  ( x3 - 1)1/3.x + x2

We obtain
                        (x3 - 1) - x3
 b = lim  ------------------------------------------------
            ( x3 - 1)2/3 +  ( x3 - 1)1/3.x + x2

                      -1
   = lim  ------------------------------------------------
           ( x3 - 1)2/3 +  ( x3 - 1)1/3.x + x2

   = 0
The oblique asymptote of g(x) is y = x.
If we shift g(x) exactly one unit to the left, then we obtain the requested asymptote y = x + 1.

-----------------

Now the alternative method.
We calculate the oblique asymptote of f(x) = (3x + 3x2 + x3)1/3 for x -> + infty.
It is the oblique asymptote of f(x)= ( (x + 1)3 - 1)1/3 for x -> + infty.
We shift the graph of f(x) one unit to the right. The new function is
g(x) = ( x3 - 1)1/3
If x -> + infty, so for very large x-values, the graphs of y =( x3 - 1)1/3 and y =( x3)1/3 are infinitely close to each other. But y = ( x3)1/3 is nothing else than the line y = x.
The line y = x is the oblique asymptote of y =( x3 - 1)1/3.
If we shift g(x) exactly one unit to the left, then we obtain the requested asymptote y = x + 1.

Example 5

Find the oblique asymtotes of y = 2x - 4 arctan(x)

We'll calculate a and b.
First x -> + infty.

 
          2x - 4 arctan(x)
a = lim --------------------
               x

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

b = lim ( -4  arctan(x) ) = -2 pi
The oblique asymtote is y = 2 x - 2 pi

Since the function y = 2x - 4 arctan(x) is odd, the graph is symmetrical to the origin (0,0).

So, the second asymtote is y = 2 x + 2 pi

Example 6

Find all asymtotes of y = arcsin(1-x2)

Since the range of the function is [-pi/2, pi/2], there are no vertical an no oblique asymtotes.

Since the domain of the function is [-sqrt(2), sqrt(2)], there are no horizontal asymtotes.





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