------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 1/24
BUILDING UP VIRTUAL MATHEMATICS LABORATORY
Partnership project LLP-2009-LEO-МP-09, MP 09-05414
Solved Problems on Limits at Infinity, Asymptotes and
Dominant terms
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 2/24
General technique
for finding limits with singularities
In all limits at infinity or at a singular finite point, where the function is undefined, we try to apply the following general technique.
This has to be known by heart:
The general technique is to isolate the singularity as a term and to try to cancel it.
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 3/24
Problem 1. Find the limit lim2 1x
xx→∞ + .
Solution. The numerator and denominator are growing to infinity at x →∞. The singular point is x=∞.
1lim lim lim1 12 1 (2 ) (2 )
1 1 11 (2 0) 2(2 lim )
→∞ →∞ →∞
→∞
= =+ + +
= = =++
x x x
x
x xx x
x x
x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 4/24
Problem 2. Find the limit 2
23 1lim
3 2x
x xx x→∞
− +− .
Solution. We apply the general technique:
22 2
2 2
2
3 1(1 )3 1lim lim 23 2 (3 )
3 1lim (1 ) 1 0 0 12 3 0 3lim (3 )
→∞ →∞
→∞
→∞
− +− +
=− −
− +− −
= = =−−
x x
x
x
xx x x xx x x
x
x x
x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 5/24
Problem 3. Find the limit 2lim
7 1x
x
x→−∞ + . Solution. The singular point is x =− ∞.
2 2 2
2 2
lim lim lim1 17 1 (7 ) (7 )
x x x
x x x
x x xx x
→−∞ →−∞ →−∞= =
+ + +
( )2 2 2
1 1lim lim1 1 1 7(7 ) 7 7 lim
x x
x
x x
x xx x x
→−∞ →−∞
→−∞
− −= = = =
+ − + +
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 6/24
Problem 4. Find the limit 1lim
1 1x x x→∞ + − − . Solution. The singular point is x = ∞. It is not difficult to observe that the denominator approaches 0 at x→∞. To isolate this singularity we rationalize the denominator by using the formula
( )( )2 2− = + −A B A B A B
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 7/24
( )( )1 1 1lim lim
1 1 1 1 1 1→∞ →∞
+ + −=
+ − − + − − + + −x x
x xx x x x x x
( ) ( ) ( )2 21 1 1 1 1 1lim lim lim
1 1 21 1→∞ →∞ →∞
+ + − + + − + + −= = =
+ − −+ − −x x x
x x x x x xx xx x
Now apply the general technique
1 1 1 1lim 1 1 lim 1 lim 1 lim2 2→∞ →∞ →∞ →∞
= + + − = + + −
x x x x
x xx x x x
( ) ( )lim 1 0 1 0 lim 1 1 lim2 2→∞ →∞ →∞
= + + − = + = = ∞x x x
x x x .
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 8/24
Problem 5. Find the asymptotes of the function
3 2( )
1+
=−
xf xx .
Solution. The domain of definition is for all 1≠x , or ( ,1) (1, )∈ −∞ ∪ ∞x .
Horizontal asymptotes at →±∞x : 2 13 3 2 lim
3 2 3 0lim lim 31 11 1 01 1 lim
→∞
→∞ →∞
→∞
+ + + + = = = =− − − −
x
x x
x
xx x xx x
x x
2 13 3 2 lim3 2 3 0lim lim 3
1 11 1 01 1 lim
→−∞
→−∞ →−∞
→−∞
+ + + + = = = =− − − −
x
x x
x
xx x xx x
x x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 9/24
We obtained that there is a horizontal asymptote 3=y . Vertical asymptotes. A finite singular point is 1=x
where the function is undefined. The left-sided limit, when 1, 1→ <x x is:
( ) [ ]11
3 2 3.1 2 5lim1 lim 1 0−
−→
→
+ += = = −∞
− − −xx
xx x
The right-sided limit, when 1, 1→ >x x is:
( ) [ ]11
3 2 5 5lim1 lim 1 0+
+→
→
+= = = ∞
− − +xx
xx x
The function has two-sided vertical asymptote 1=x .
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 10/24
Oblique asymptotes. We look for the existence of two numbers k and b, such that at →∞x
( )lim
→∞=
x
f xkx and ( )lim ( )
→∞= −
xb f x kx (or at →−∞x ).
Then the line = +y kx b is an oblique asymptote of f (x).
( )( )( )
( )( )
23 2 /( ) 3 2lim lim lim
1 1 1/
3 2 /1lim . lim 01 1/
→∞ →∞ →∞
→∞ →∞
++= = =
− −
+= =
−
x x x
x x
x xf x xkx x x x x
xx x
The same is for →−∞x . Therefore there is no an oblique asymptote.
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 11/24
Problem 6. Draw the graph of the function
3 2( )
1+
=−
xf xx .
Using the asymptotes from Problem 5 we obtain the graphics in Fig. 1.
x
y
1
f (x)
y = 3
x=110 5 5 10
2
2
4
6
8
Fig. 1 Graphics of the function 3 2( )
1+
=−
xf xx with its
asymptotes (in blue color).
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 12/24
Problem 7. Find the asymptotes of the function 2( ) 2 4 25= − −f x x x .
Solution. The domain of definition is 24 25 0− ≥x , or ( , 2.5] [ 2.5, )∈ −∞ ∪ ∞x .
First look for horizontal asymptotes. At →∞x :
( ) ( )( )( )
2 22
2
2 4 25 2 4 25lim 2 4 25 lim
2 4 25→∞ →∞
− − + −− − =
+ −x x
x x x xx x
x x
( ) ( )2 2
2 2
4 4 25 25lim lim2 4 25 2 1 1 25 / 4→∞ →∞
− += =
+ − + −x x
x x
x x x x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 13/24
2
25 1 1 25 1 1 25 1lim . lim lim . lim 02 2 41 1 01 1 25 / 4→∞ →∞ →∞ →∞
= = = =+ −+ −x x x xx x xx
There is a →∞xhorizontal asymptote y = 0 at . At →−∞x :
( )2 22
25lim 2 4 25 lim 2 4 14→−∞ →−∞
− − = − − x xx x x x
x
2 225 25lim 2 2 1 lim 2 2 1
4 4→−∞ →−∞
= − − = + − =
x xx x x x
x x
( )225lim 2 . 1 1 lim lim 2 . 1 1 0 4. lim
4→−∞ →−∞ →−∞ →−∞
= + − = + − = = −∞
x x x xx x x
x→−∞xThere is no horizontal asymptote at .
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 14/24
There is no finite singular points and no vertical asymptotes.
We look for oblique asymptotes at →−∞x
2 2252 2 1
( ) 2 4 25 4lim lim lim→−∞ →−∞ →−∞
− −− −
= = =x x x
x xf x x x xk
x x x
22
2525 2 1 12 2 144lim lim
→−∞ →−∞
+ − + −
= = =x x
xx xxx
x x
2252 lim 1 1 2.2 4
4→−∞
= + − = =
x x ⇒ k = 4.
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 15/24
( ) ( )2lim ( ) lim 2 4 25 4→−∞ →−∞
= − = − − −x x
b f x kx x x x
( )2lim 2 4 25→−∞
= − − −x
x x . As there is [∞−∞ ] we will rationalize the numerator
( )( )( )
( )( )
2 2 2 2
2 2
2 4 25 2 4 25 4 25 4lim lim
2 4 25 2 4 25→−∞ →−∞
− − − − + − − −=
− + − − + −x x
x x x x x x
x x x x
2
25 25 25lim lim lim 042 2 1 02 2 1 25 /→−∞ →−∞ →−∞
− − −= = =
−− − −− + −x x x xx xx x x ⇒ b = 0.
We obtained the oblique asymptote y = 4x at →−∞x .
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 16/24
Problem 8. Draw the graph of the function
2( ) 2 4 25= − −f x x x .
We find (2.5) 5, ( 2.5) 5= − =−f f .
Using the asymptotes from Problem 5 we obtain the graphics in Fig. 2.
2.5-2.5
f (x)
f (x)
y = 0
y = 4x
x
y
10 5 5 10
40
30
20
10
Fig. 2 Graphics of the function ( )2( ) 2 4 25= − −f x x x with its asymptotes (in blue color).
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 17/24
Problem 9. Find the asymptotes and draw graphics of the function
3 2
22 4 15( )
3 12− +
=−
x xf xx
Solution. The domain of definition is: 23 12 0− ≠x , or
( , 2) ( 2, 2) (2, )∈ −∞ − ∪ − ∪ ∞x . The singular points of the function are 1 22, 2= − =x x .
Horizontal asymptotes. We compute the limits at infinities:
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 18/24
( )( )
3 23 2
2 2 2
2 4 / 15 /2 4 15 2lim lim lim33 12 3 12 /→±∞ →±∞ →±∞
− +− += = = ±∞
− −x x x
x x xx x xx x x .
This means the absence of the horizontal asymptotes.
Vertical asymptotes. We find the limits at the singular points:
… [ ]3 2
22 2
2 4 15 15lim lim03 12+ +→ →
− += = ∞
+−x x
x xx
… [ ]3 2
22 2
2 4 15 15lim lim03 12− −→ →
− += = −∞
−−x x
x xx
… [ ]3 2
22 2
2 4 15 17lim lim03 12+ +→− →
− + −= = −∞
+−x x
x xx
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 19/24
… [ ]3 2
22 2
2 4 15 17lim lim03 12− −→− →−
− + −= = ∞
−−x x
x xx
We can conclude that the lines 2, 2= − =x x are two sided vertical asymptotes. Oblique asymptotes. For both ±∞
⇒ 23
=k .
( )( )
( )
3 2
2
3 2
3 2
( ) 2 4 15lim lim3 12
2 4 / 15 / 2lim33 12 /
→±∞ →±∞
→±∞
− += =
−
− += =
−
x x
x
f x x xkx x x
x x x
x x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 20/24
( )3 2
22 4 15 2lim ( ) lim
33 12→±∞ →±∞
− += − = − − x x
x xb f x kx xx
3 2 3 2
2 22 4 15 2 8 4 8 15 4lim lim
33 12 3 12→±∞ →±∞
− + − + − + += = = − − − x x
x x x x x xx x
⇒ 43
= −b .
The line 2 43 3
= −y x is an oblique asymptote at infinity and negative infinity.
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 21/24
Representation of the graphics of the function
3 2
22 4 15( )
3 12− +
=−
x xf xx
and its asymptotes: x = ±2 , vertical two- sided asymptotes; y = 2/3x -4/3, oblique asymptote at x→±∞, see Fig. 3.
x
y
f (x)
x = 2
x = - 2
y = 2/3x - 4/3
2-210 5 5 10
20
10
10
20
Fig. 3 Graphics of the function 3 2
22 4 15( )
3 12− +
=−
x xf xx (red
color) with its asymptotes (blue color).
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 22/24
Problem 10. Find the dominant terms of the function 3 2
22 4 3( )
1− +
=+
x xf xx
Solution. We divide the polynomials of the numerator and denominator to obtain the entire part and the remainder:
23 22
3
2
2
2 7: 1 2 42 4 31
2 2
4 2 3
4 42 7
− ++ = − +− +
++
− − +
− −− +
xx xx xx
x x
x x
xx
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 23/24
Here the term 2 4−x is the entire part and 2 7− +x is the remainder. This way we obtained the representation of the function with its dominant terms:
3 2
2 22 4 3 2 7( ) 2 4
1 1− + − +
= = − ++ +
x x xf x xx x .
Having the dominant terms it is easy to observe, that ( ) 2 4≈ −f x x for large x, and lim ( )
→±∞= ±∞
xf x
------------------ Snezhana Gocheva-Ilieva, Plovdiv University --------------------- 24/24
Further reading:
[1] G. B. Thomas, M. D. Weir., J. Hass, F. R. Giordano, Thomas’ Calculus including second-order differential equations, 11 ed., Pearson Addison-Wesley, 2005.
[2] http://www.wolframalpha.com/
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