EVALUATING LIMITS AND FIND HORIZONTAL AND VERTICAL ASYMPTOTES

A function f(x) will have the horizontal asymptote y = L if either

lim x→∞  f(x) = L or lim x→−∞ f(x) = L

To find horizontal asymptotes, we simply evaluate the limit of the function as it approaches infinity, and again as it approaches negative infinity.

Evaluate each limit and then identify any horizontal asymptotes.

Problem 1 :

xx3+8x-42x3+3

Solution:

=xx3+8x-42x3+3=x1+8x2-4x32+3x3=1+82-432+33=1+0-02+0=12

Horizontal asymptote:

Degree of numerator = 3

Degree of denominator = 3

degree of numerator = degree of denominator

y = leading coefficient of N(x) / coefficient of D(x)

y = 1/2

So, equation of the horizontal asymptote is y = 1/2.

Problem 2 :

xx2-12x3-8x2+3

Solution:

=xx2-12x3-8x2+3=x1x-1x32-8x+3x3=1-132-8+33=02=0

Horizontal asymptote:

Degree of numerator = 2

Degree of denominator = 3

degree of numerator < degree of denominator

So, equation of the horizontal asymptote is y = 0 which is the x-axis.

Problem 3 :

xx5-x-16x3

Solution:

=xx5-x-16x3=x1-1x4-1x56x2=1-14-1562=10=

Horizontal asymptote:

Degree of numerator = 5

Degree of denominator = 3

degree of numerator > degree of denominator

So, there is no horizontal asymptote. Slant  asymptote  will be there.

Problem 4 :

xx+1x2-1

Solution:

=xx+1x2-1=x1x+1x21-1x2=1+121-12=01=0

Horizontal asymptote:

Degree of numerator = 1

Degree of denominator = 2

degree of numerator < degree of denominator

So, equation of the horizontal asymptote is y = 0 which is the x-axis.

Problem 5 :

xx2-1x+1

Solution:

=xx2-1x+1=x(x+1)(x-1)(x+1)=x(x-1)=-1=

Horizontal asymptote:

Degree of numerator = 2

Degree of denominator = 1

degree of numerator > degree of denominator

So, there is no horizontal asymptote.

Evaluate each limit and then identify any vertical asymptotes.

Problem 6 :

x25x2+xx-2

Solution:

=x25x2+xx-2=5(2)2+22-2=220=undefined

Vertical asymptote:

x25x2+xx-2

Equate the denominator to zero and solve for x.

x - 2 = 0

x = 2

So, the equation of the vertical asymptote is x = 2.

Problem 7 :

x1x(x-1)3

Solution:

=x1x(x-1)3=1(1-1)3=10=undefined

Vertical asymptote:

x1x(x-1)3

Equate the denominator to zero and solve for x.

(x - 1)3 = 0

x = 1

So, the equation of the vertical asymptote is x = 1.

Problem 8 :

x-3x2+2x-15x2-9

Solution:

=x-3x2+2x-15x2-9=(-3)2+2(-3)-15(-3)2-9=-120=undefined

Vertical asymptote:

x-3x2+2x-15x2-9

Equate the denominator to zero and solve for x.

x2 - 9 = 0

x2 = 9

x = 3

So, the equation of the vertical asymptote is x = 3.

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More