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Evaluate lim x 0 + x x .

1

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Growth rates of functions

Suppose the functions f and g both approach infinity as x . Although the values of both functions become arbitrarily large as the values of x become sufficiently large, sometimes one function is growing more quickly than the other. For example, f ( x ) = x 2 and g ( x ) = x 3 both approach infinity as x . However, as shown in the following table, the values of x 3 are growing much faster than the values of x 2 .

Comparing the growth rates of x 2 And x 3
x 10 100 1000 10,000
f ( x ) = x 2 100 10,000 1,000,000 100,000,000
g ( x ) = x 3 1000 1,000,000 1,000,000,000 1,000,000,000,000

In fact,

lim x x 3 x 2 = lim x x = . or, equivalently, lim x x 2 x 3 = lim x 1 x = 0 .

As a result, we say x 3 is growing more rapidly than x 2 as x . On the other hand, for f ( x ) = x 2 and g ( x ) = 3 x 2 + 4 x + 1 , although the values of g ( x ) are always greater than the values of f ( x ) for x > 0 , each value of g ( x ) is roughly three times the corresponding value of f ( x ) as x , as shown in the following table. In fact,

lim x x 2 3 x 2 + 4 x + 1 = 1 3 .
Comparing the growth rates of x 2 And 3 x 2 + 4 x + 1
x 10 100 1000 10,000
f ( x ) = x 2 100 10,000 1,000,000 100,000,000
g ( x ) = 3 x 2 + 4 x + 1 341 30,401 3,004,001 300,040,001

In this case, we say that x 2 and 3 x 2 + 4 x + 1 are growing at the same rate as x .

More generally, suppose f and g are two functions that approach infinity as x . We say g grows more rapidly than f as x if

lim x g ( x ) f ( x ) = ; or, equivalently, lim x f ( x ) g ( x ) = 0 .

On the other hand, if there exists a constant M 0 such that

lim x f ( x ) g ( x ) = M ,

we say f and g grow at the same rate as x .

Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.

Comparing the growth rates of ln ( x ) , x 2 , And e x

For each of the following pairs of functions, use L’Hôpital’s rule to evaluate lim x ( f ( x ) g ( x ) ) .

  1. f ( x ) = x 2 and g ( x ) = e x
  2. f ( x ) = ln ( x ) and g ( x ) = x 2
  1. Since lim x x 2 = and lim x e x , we can use L’Hôpital’s rule to evaluate lim x [ x 2 e x ] . We obtain
    lim x x 2 e x = lim x 2 x e x .

    Since lim x 2 x = and lim x e x = , we can apply L’Hôpital’s rule again. Since
    lim x 2 x e x = lim x 2 e x = 0 ,

    we conclude that
    lim x x 2 e x = 0 .

    Therefore, e x grows more rapidly than x 2 as x (See [link] and [link] ).
    The functions g(x) = ex and f(x) = x2 are graphed. It is obvious that g(x) increases much more quickly than f(x).
    An exponential function grows at a faster rate than a power function.
    Growth rates of a power function and an exponential function.
    x 5 10 15 20
    x 2 25 100 225 400
    e x 148 22,026 3,269,017 485,165,195
  2. Since lim x ln x = and lim x x 2 = , we can use L’Hôpital’s rule to evaluate lim x ln x x 2 . We obtain
    lim x ln x x 2 = lim x 1 / x 2 x = lim x 1 2 x 2 = 0 .

    Thus, x 2 grows more rapidly than ln x as x (see [link] and [link] ).
    The functions g(x) = x2 and f(x) = ln(x) are graphed. It is obvious that g(x) increases much more quickly than f(x).
    A power function grows at a faster rate than a logarithmic function.
    Growth rates of a power function and a logarithmic function
    x 10 100 1000 10,000
    ln ( x ) 2.303 4.605 6.908 9.210
    x 2 100 10,000 1,000,000 100,000,000
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Compare the growth rates of x 100 and 2 x .

The function 2 x grows faster than x 100 .

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Using the same ideas as in [link] a. it is not difficult to show that e x grows more rapidly than x p for any p > 0 . In [link] and [link] , we compare e x with x 3 and x 4 as x .

This figure has two figures marked a and b. In figure a, the functions y = ex and y = x3 are graphed. It is obvious that ex increases more quickly than x3. In figure b, the functions y = ex and y = x4 are graphed. It is obvious that ex increases much more quickly than x4, but the point at which that happens is further to the right than it was for x3.
The exponential function e x grows faster than x p for any p > 0 . (a) A comparison of e x with x 3 . (b) A comparison of e x with x 4 .
An exponential function grows at a faster rate than any power function
x 5 10 15 20
x 3 125 1000 3375 8000
x 4 625 10,000 50,625 160,000
e x 148 22,026 3,269,017 485,165,195

Similarly, it is not difficult to show that x p grows more rapidly than ln x for any p > 0 . In [link] and [link] , we compare ln x with x 3 and x .

This figure shows y = the square root of x, y = the cube root of x, and y = ln(x). It is apparent that y = ln(x) grows more slowly than either of these functions.
The function y = ln ( x ) grows more slowly than x p for any p > 0 as x .
A logarithmic function grows at a slower rate than any root function
x 10 100 1000 10,000
ln ( x ) 2.303 4.605 6.908 9.210
x 3 2.154 4.642 10 21.544
x 3.162 10 31.623 100

Key concepts

  • L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form 0 0 or / arises.
  • L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form 0 0 or / .
  • The exponential function e x grows faster than any power function x p , p > 0 .
  • The logarithmic function ln x grows more slowly than any power function x p , p > 0 .

For the following exercises, evaluate the limit.

Evaluate the limit lim x e x x .

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Evaluate the limit lim x e x x k .

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Evaluate the limit lim x ln x x k .

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Evaluate the limit lim x a x a x 2 a 2 .

1 2 a

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Evaluate the limit lim x a x a x 3 a 3 .

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Evaluate the limit lim x a x a x n a n .

1 n a n 1

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For the following exercises, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule.

lim x x 1 / x

Cannot apply directly; use logarithms

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lim x 0 x 2 1 / x

Cannot apply directly; rewrite as lim x 0 x 3

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For the following exercises, evaluate the limits with either L’Hôpital’s rule or previously learned methods.

lim x 3 x 2 9 x 3

6

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lim x 3 x 2 9 x + 3

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lim x 0 ( 1 + x ) −2 1 x

−2

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lim x π / 2 cos x 2 π x

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lim x π x π sin x

−1

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lim x 1 x 1 sin x

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lim x 0 ( 1 + x ) n 1 x

n

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lim x 0 ( 1 + x ) n 1 n x x 2

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lim x 0 sin x tan x x 3

1 2

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lim x 0 1 + x 1 x x

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lim x 0 e x x 1 x 2

1 2

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lim x 1 x 1 ln x

1

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lim x 0 ( x + 1 ) 1 / x

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lim x 1 x x 3 x 1

1 6

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lim x x sin ( 1 x )

1

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lim x 0 sin x x x 2

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lim x 0 + x ln ( x 4 )

0

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lim x ( x e x )

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lim x x 2 e x

0

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lim x 0 3 x 2 x x

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lim x 0 1 + 1 / x 1 1 / x

−1

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lim x π / 4 ( 1 tan x ) cot x

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lim x x e 1 / x

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lim x 0 ( 1 1 x ) x

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lim x ( 1 1 x ) x

1 e

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For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly.

[T] lim x 0 e x 1 x

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[T] lim x 0 x sin ( 1 x )

0

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[T] lim x 1 x 1 1 cos ( π x )

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[T] lim x 1 e ( x 1 ) 1 x 1

1

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[T] lim x 1 ( x 1 ) 2 ln x

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[T] lim x π 1 + cos x sin x

0

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[T] lim x 0 ( csc x 1 x )

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[T] lim x 0 + tan ( x x )

tan ( 1 )

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[T] lim x 0 + ln x sin x

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[T] lim x 0 e x e x x

2

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Practice Key Terms 2

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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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