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
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.
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
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.
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
|
f(x)
lim ---- = O
infty x
There is no oblique asymptote for f(x).
f(x)
lim ---- = O
infty x
There is no oblique asymptote for f(x).
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
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. |
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.
________
| 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.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.
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.
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
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.-----------------
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.
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
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.