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Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.

End behavior for polynomial functions

Consider the power function f ( x ) = x n where n is a positive integer. From [link] and [link] , we see that

lim x x n = ; n = 1 , 2 , 3 ,…

and

lim x x n = { ; n = 2 , 4 , 6 ,… ; n = 1 , 3 , 5 ,… .
The functions x2, x4, and x6 are graphed, and it is apparent that as the exponent grows the functions increase more quickly.
For power functions with an even power of n , lim x x n = = lim x x n .
The functions x, x3, and x5 are graphed, and it is apparent that as the exponent grows the functions increase more quickly.
For power functions with an odd power of n , lim x x n = and lim x x n = .

Using these facts, it is not difficult to evaluate lim x c x n and lim x c x n , where c is any constant and n is a positive integer. If c > 0 , the graph of y = c x n is a vertical stretch or compression of y = x n , and therefore

lim x c x n = lim x x n and lim x c x n = lim x x n if c > 0 .

If c < 0 , the graph of y = c x n is a vertical stretch or compression combined with a reflection about the x -axis, and therefore

lim x c x n = lim x x n and lim x c x n = lim x x n if c < 0 .

If c = 0 , y = c x n = 0 , in which case lim x c x n = 0 = lim x c x n .

Limits at infinity for power functions

For each function f , evaluate lim x f ( x ) and lim x f ( x ) .

  1. f ( x ) = −5 x 3
  2. f ( x ) = 2 x 4
  1. Since the coefficient of x 3 is −5 , the graph of f ( x ) = −5 x 3 involves a vertical stretch and reflection of the graph of y = x 3 about the x -axis. Therefore, lim x ( −5 x 3 ) = and lim x ( −5 x 3 ) = .
  2. Since the coefficient of x 4 is 2 , the graph of f ( x ) = 2 x 4 is a vertical stretch of the graph of y = x 4 . Therefore, lim x 2 x 4 = and lim x 2 x 4 = .
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Let f ( x ) = −3 x 4 . Find lim x f ( x ) .

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We now look at how the limits at infinity for power functions can be used to determine lim x ± f ( x ) for any polynomial function f . Consider a polynomial function

f ( x ) = a n x n + a n 1 x n 1 + + a 1 x + a 0

of degree n 1 so that a n 0 . Factoring, we see that

f ( x ) = a n x n ( 1 + a n 1 a n 1 x + + a 1 a n 1 x n 1 + a 0 a n ) .

As x ± , all the terms inside the parentheses approach zero except the first term. We conclude that

lim x ± f ( x ) = lim x ± a n x n .

For example, the function f ( x ) = 5 x 3 3 x 2 + 4 behaves like g ( x ) = 5 x 3 as x ± as shown in [link] and [link] .

Both functions f(x) = 5x3 – 3x2 + 4 and g(x) = 5x3 are plotted. Their behavior for large positive and large negative numbers converges.
The end behavior of a polynomial is determined by the behavior of the term with the largest exponent.
A polynomial’s end behavior is determined by the term with the largest exponent.
x 10 100 1000
f ( x ) = 5 x 3 3 x 2 + 4 4704 4,970,004 4,997,000,004
g ( x ) = 5 x 3 5000 5,000,000 5,000,000,000
x −10 −100 −1000
f ( x ) = 5 x 3 3 x 2 + 4 −5296 −5,029,996 −5,002,999,996
g ( x ) = 5 x 3 −5000 −5,000,000 −5,000,000,000

End behavior for algebraic functions

The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In [link] , we show that the limits at infinity of a rational function f ( x ) = p ( x ) q ( x ) depend on the relationship between the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of x appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of x .

Determining end behavior for rational functions

For each of the following functions, determine the limits as x and x . Then, use this information to describe the end behavior of the function.

  1. f ( x ) = 3 x 1 2 x + 5 (Note: The degree of the numerator and the denominator are the same.)
  2. f ( x ) = 3 x 2 + 2 x 4 x 3 5 x + 7 (Note: The degree of numerator is less than the degree of the denominator.)
  3. f ( x ) = 3 x 2 + 4 x x + 2 (Note: The degree of numerator is greater than the degree of the denominator.)
  1. The highest power of x in the denominator is x . Therefore, dividing the numerator and denominator by x and applying the algebraic limit laws, we see that
    lim x ± 3 x 1 2 x + 5 = lim x ± 3 1 / x 2 + 5 / x = lim x ± ( 3 1 / x ) lim x ± ( 2 + 5 / x ) = lim x ± 3 lim x ± 1 / x lim x ± 2 + lim x ± 5 / x = 3 0 2 + 0 = 3 2 .

    Since lim x ± f ( x ) = 3 2 , we know that y = 3 2 is a horizontal asymptote for this function as shown in the following graph.
    The function f(x) = (3x + 1)/(2x + 5) is plotted as is its horizontal asymptote at y = 3/2.
    The graph of this rational function approaches a horizontal asymptote as x ± .
  2. Since the largest power of x appearing in the denominator is x 3 , divide the numerator and denominator by x 3 . After doing so and applying algebraic limit laws, we obtain
    lim x ± 3 x 2 + 2 x 4 x 3 5 x + 7 = lim x ± 3 / x + 2 / x 2 4 5 / x 2 + 7 / x 3 = 3.0 + 2.0 4 5.0 + 7.0 = 0 .

    Therefore f has a horizontal asymptote of y = 0 as shown in the following graph.
    The function f(x) = (3x2 + 2x)/(4x2 – 5x + 7) is plotted as is its horizontal asymptote at y = 0.
    The graph of this rational function approaches the horizontal asymptote y = 0 as x ± .
  3. Dividing the numerator and denominator by x , we have
    lim x ± 3 x 2 + 4 x x + 2 = lim x ± 3 x + 4 1 + 2 / x .

    As x ± , the denominator approaches 1 . As x , the numerator approaches + . As x , the numerator approaches . Therefore lim x f ( x ) = , whereas lim x f ( x ) = as shown in the following figure.
    The function f(x) = (3x2 + 4x)/(x + 2) is plotted. It appears to have a diagonal asymptote as well as a vertical asymptote at x = −2.
    As x , the values f ( x ) . As x , the values f ( x ) .
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Questions & Answers

f(x) =3+8+4
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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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