Solved Problems on Limits at Infinity, Asymptotes and ... Solved Problems... · In all limits at infinity or at a singular finite point, where the function is undefined, we try to

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Solved Problems on Limits at Infinity, Asymptotes and

Dominant terms


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.


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


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


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

→−∞ →−∞

→−∞

− −= = = =

+ − + +


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


( )( )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 .


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


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 .


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.


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).


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


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 .


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.


( ) ( )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 .


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).


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:


( )( )

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


… [ ]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


( )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.


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).


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


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


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/

http://www.wolframalpha.com/�

Solved Problems on Limits at Infinity, Asymptotes and ... Solved Problems... · In all limits at infinity or at a singular finite point, where the function is undefined, we try to

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