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Consider the function f ( x ) = 5 x 2 / 3 . Determine the point on the graph where a cusp is located. Determine the end behavior of f .

The function f has a cusp at ( 0 , 5 ) lim x 0 f ( x ) = , lim x 0 + f ( x ) = . For end behavior, lim x ± f ( x ) = .

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Key concepts

  • The limit of f ( x ) is L as x (or as x ) if the values f ( x ) become arbitrarily close to L as x becomes sufficiently large.
  • The limit of f ( x ) is as x if f ( x ) becomes arbitrarily large as x becomes sufficiently large. The limit of f ( x ) is as x if f ( x ) < 0 and | f ( x ) | becomes arbitrarily large as x becomes sufficiently large. We can define the limit of f ( x ) as x approaches similarly.
  • For a polynomial function p ( x ) = a n x n + a n 1 x n 1 + + a 1 x + a 0 , where a n 0 , the end behavior is determined by the leading term a n x n . If n 0 , p ( x ) approaches or at each end.
  • For a rational function f ( x ) = p ( x ) q ( x ) , the end behavior is determined by the relationship between the degree of p and the degree of q . If the degree of p is less than the degree of q , the line y = 0 is a horizontal asymptote for f . If the degree of p is equal to the degree of q , then the line y = a n b n is a horizontal asymptote, where a n and b n are the leading coefficients of p and q , respectively. If the degree of p is greater than the degree of q , then f approaches or at each end.

For the following exercises, examine the graphs. Identify where the vertical asymptotes are located.

For the following functions f ( x ) , determine whether there is an asymptote at x = a . Justify your answer without graphing on a calculator.

f ( x ) = x + 1 x 2 + 5 x + 4 , a = −1

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f ( x ) = x x 2 , a = 2

Yes, there is a vertical asymptote

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f ( x ) = ( x + 2 ) 3 / 2 , a = −2

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f ( x ) = ( x 1 ) −1 / 3 , a = 1

Yes, there is vertical asymptote

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f ( x ) = 1 + x −2 / 5 , a = 1

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For the following exercises, evaluate the limit.

lim x 1 3 x + 6

0

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lim x 2 x 5 4 x

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lim x x 2 2 x + 5 x + 2

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lim x 3 x 3 2 x x 2 + 2 x + 8

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lim x x 4 4 x 3 + 1 2 2 x 2 7 x 4

1 7

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lim x 3 x x 2 + 1

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lim x 4 x 2 1 x + 2

−2

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lim x 4 x x 2 1

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lim x 4 x x 2 1

−4

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lim x 2 x x x + 1

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For the following exercises, find the horizontal and vertical asymptotes.

f ( x ) = x 9 x

Horizontal: none, vertical: x = 0

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f ( x ) = x 3 4 x 2

Horizontal: none, vertical: x = ± 2

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f ( x ) = x 2 + 3 x 2 + 1

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f ( x ) = sin ( x ) sin ( 2 x )

Horizontal: none, vertical: none

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f ( x ) = cos x + cos ( 3 x ) + cos ( 5 x )

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f ( x ) = x sin ( x ) x 2 1

Horizontal: y = 0 , vertical: x = ± 1

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f ( x ) = 1 x 3 + x 2

Horizontal: y = 0 , vertical: x = 0 and x = −1

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f ( x ) = 1 x 1 2 x

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f ( x ) = x 3 + 1 x 3 1

Horizontal: y = 1 , vertical: x = 1

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f ( x ) = sin x + cos x sin x cos x

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f ( x ) = x sin x

Horizontal: none, vertical: none

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For the following exercises, construct a function f ( x ) that has the given asymptotes.

x = 1 and y = 2

Answers will vary, for example: y = 2 x x 1

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y = 4 , x = −1

Answers will vary, for example: y = 4 x x + 1

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For the following exercises, graph the function on a graphing calculator on the window x = [ −5 , 5 ] and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.

[T] f ( x ) = 1 x + 10

y = 0

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[T] f ( x ) = x + 1 x 2 + 7 x + 6

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[T] lim x x 2 + 10 x + 25

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[T] lim x x + 2 x 2 + 7 x + 6

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[T] lim x 3 x + 2 x + 5

y = 3

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For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior.

y = 2 x + 1 x 2 + 6 x + 5

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y = x 3 + 4 x 2 + 3 x 3 x + 9


An upward-facing parabola with minimum between x = 0 and x = −1 with y intercept between 0 and 1.

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y = x 2 + x 2 x 2 3 x 4

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y = cos x x , on x = [ −2 π , 2 π ]


This graph has vertical asymptote at x = 0. The first part of the function occurs in the second and third quadrants and starts in the third quadrant just below (−2π, 0), increases and passes through the x axis at −3π/2, reaches a maximum and then decreases through the x axis at −π/2 before approaching the asymptote. On the other side of the asymptote, the function starts in the first quadrant, decreases quickly to pass through π/2, decreases to a local minimum and then increases through (3π/2, 0) before staying just above (2π, 0).

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y = x tan x , x = [ π , π ]


This graph has vertical asymptotes at x = ±π/2. The graph is symmetric about the y axis, so describing the left hand side will be sufficient. The function starts at (−π, 0) and decreases quickly to the asymptote. Then it starts on the other side of the asymptote in the second quadrant and decreases to the the origin.

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y = x 2 sin ( x ) , x = [ −2 π , 2 π ]


This function starts at (−2π, 0), increases to near (−3π/2, 25), decreases through (−π, 0), achieves a local minimum and then increases through the origin. On the other side of the origin, the graph is the same but flipped, that is, it is congruent to the other half by a rotation of 180 degrees.

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For f ( x ) = P ( x ) Q ( x ) to have an asymptote at y = 2 then the polynomials P ( x ) and Q ( x ) must have what relation?

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For f ( x ) = P ( x ) Q ( x ) to have an asymptote at x = 0 , then the polynomials P ( x ) and Q ( x ) . must have what relation?

Q ( x ) . must have have x k + 1 as a factor, where P ( x ) has x k as a factor.

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If f ( x ) has asymptotes at y = 3 and x = 1 , then f ( x ) has what asymptotes?

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Both f ( x ) = 1 ( x 1 ) and g ( x ) = 1 ( x 1 ) 2 have asymptotes at x = 1 and y = 0 . What is the most obvious difference between these two functions?

lim x 1 f ( x ) and lim x 1 g ( x )

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True or false: Every ratio of polynomials has vertical asymptotes.

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Practice Key Terms 5

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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