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
3x^{2} - 4x -1 lim -------------- = 0.5 infty 6x^{2} - 6 3x^{2} - 4x -1 So, y = 0.5 is a horizontal asymptote of a function ------------- 6x^{2} - 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-> inftyFrom (*) 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 xWith (*) we'll construct a formula for b.
(*) => lim (f(x) - ax ) = b (2) inftySince 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 xThere is no oblique asymptote for f(x).
f(x) lim ---- = O infty xThere is no oblique asymptote for f(x).
f(x) a = lim ---- = a finite number , not 0 infty xThere is an oblique asymptote for f(x).
Concrete examples:
3x^{3} + 2x +1 ------------- has no oblique asymptote 6 x + 5 3x^{3} + 2x +1 --------------- has no oblique asymptote 6 x^{3} + 5x 3x^{3} + 2x +1 ------------- has an oblique asymptote 6 x^{2} + 5 We calculate oblique asymptote 3x^{3} + 2x +1 a = lim ------------------ = 1/2 infty 6 x^{3} + 5x 3 x^{3} + 2x +1 x b = lim ( ----------------- - ----- ) infty 6 x^{2} + 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
3x^{3} + 2x +1 y = ------------- 6 x + 5 The quotient of the division of x^{3} + 2x +1 by 6 x + 5 is x/2 The oblique asymptote is x/2General 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 x^{4} - 3 x^{2} + x f(x) = -------------------- 3 x^{3} - 2x^{2} + 4The 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.Alternative method:
First, let x -> + infty . Then ________ | 2 \| x - 1 + 2 a = lim ------------------ x __________ | \| 1 - 1/x^{2} + 2/x = lim ----------------------- = 1 1 ________ | 2 b = lim \| x - 1 + 2 - 1.x (sqrt(x^{2} -1) - x )( sqrt(x^{2} -1) + x) = lim --------------------------------------- + 2 sqrt(x^{2} -1) + x x^{2} - 1 - x^{2} = lim ----------------------------- + 2 sqrt(x^{2} -1) + x - 1 = lim --------------------- + 2 sqrt(x^{2} -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/x^{2} + 2/x = lim ----------------------- = -1 1 ________ | 2 b = lim \| x - 1 + 2 + 1.x (sqrt(x^{2} -1) + x )( sqrt(x^{2} -1) - x) = lim --------------------------------------- + 2 sqrt(x^{2} -1) - x x^{2} - 1 - x^{2} = lim ----------------------------- + 2 sqrt(x^{2} -1) - x -1 = lim --------------------- + 2 sqrt(x^{2} -1) + x = 2 So the asymptote is y = - x + 2 if x -> - infty
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 + 2are 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 + 2are 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) =
x^{2} + x -2 ----------------- sqrt(x^{2} - x - 2)The asymptote has the form y = ax+b. We calculate a and b.
x^{2} + x -2 a = lim ------------------- x sqrt(x^{2} - x - 2) 1 + 1/x - 2/x^{2} = lim ---------------------- - sqrt(1 -1/x - 2/x^{2}) = -1 x^{2} + x -2 b = lim (------------------- + x) sqrt(x^{2} - x - 2) x^{2} + x -2 + x sqrt(x^{2} - x - 2) = lim --------------------------------- sqrt(x^{2} - x - 2) ( x^{2} + x -2 + x sqrt(x^{2} - x - 2)) ( x^{2} + x -2 - x sqrt(x^{2} - x - 2)) = lim -------------------------------------------------------------------------- sqrt(x^{2} - x - 2) ( x^{2} + x -2 - x sqrt(x^{2} - x - 2)) (x^{2} + x -2)^{2} - x^{2} (x^{2} - x - 2) = lim ---------------------------------------------------------- sqrt(x^{2} - x - 2) ( x^{2} + x -2 - x sqrt(x^{2} - x - 2)) 3 x^{3} - x^{2} - 4x -4 = lim ---------------------------------------------------------- sqrt(x^{2} - x - 2) ( x^{2} + x -2 - x sqrt(x^{2} - x - 2)) We divide numerator and denominator by x^{3} 3 - 1/x - 4/x^{2} - 4/x^{3} = lim -------------------------------------------------------------- -sqrt(1 - 1/x - 2/x^{2}) (1 + 1/x -2/x^{2} +sqrt( 1 - 1/x -2/x^{2})) = 3/(-2) The oblique asymptote is y = - x -3/2
f(x) = (3x + 3x^{2} + x^{3})^{1/3}
We know 1 + 3x + 3x^{2} + x^{3} = (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) = ( x^{3} - 1)^{1/3}
We calculate the oblique asymptote y = ax + b of g(x) for x -> + infty.
( x^{3} - 1)^{1/3} a = lim ------------- x x^{3} - 1 = lim (--------- )^{1/3} x^{3} 1 = lim ( 1 - ---- )^{1/3} x^{3} = 1 b = lim ( ( x^{3} - 1)^{1/3} - x ) We multiply numerator and denominator with ( x^{3} - 1)^{2/3} + ( x^{3} - 1)^{1/3}.x + x^{2} We obtain (x^{3} - 1) - x^{3} b = lim ------------------------------------------------ ( x^{3} - 1)^{2/3} + ( x^{3} - 1)^{1/3}.x + x^{2} -1 = lim ------------------------------------------------ ( x^{3} - 1)^{2/3} + ( x^{3} - 1)^{1/3}.x + x^{2} = 0The oblique asymptote of g(x) is y = x.
-----------------
Now the alternative method.
We calculate the oblique asymptote of f(x) = (3x + 3x^{2} + x^{3})^{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) = ( x^{3} - 1)^{1/3}
If x -> + infty, so for very large x-values, the graphs of y =( x^{3} - 1)^{1/3} and
y =( x^{3})^{1/3} are infinitely close to each other.
But y = ( x^{3})^{1/3} is nothing else than the line y = x.
The line y = x is the oblique asymptote of y =( x^{3} - 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 piThe 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.