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Many times a simple algebraic manipulation involving fractions will put a limit into a form which can be studied using l H pital s Rule EXAMPLE 510
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Evaluate the limit
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x 0
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1 sin 3x
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SOLUTION We put the fractions over a common denominator to rewrite our limit as lim 3x sin 3x 3x sin 3x
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Both numerator and denominator vanish as x 0 Thus the quotient has indeterminate form 0/0 By l H pital s Rule, the limit is therefore equal to
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3 3 cos 3x 3 sin 3x + 9x cos 3x
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This quotient is still indeterminate; we apply l H pital s Rule again to obtain 9 sin 3x = 0 x 0 18 cos 3x 27x sin 3x lim EXAMPLE 511
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Evaluate the limit
x 0
1 4x
1 e4x
CHAPTER 5 Indeterminate Forms
SOLUTION The expression is indeterminate of the form We put the two fractions over a common denominator to obtain e4x 1 4x x 0 4x(e4x 1) lim Notice that the numerator and denominator both tend to zero as x 0, so this is indeterminate of the form 0/0 Therefore l H pital s Rule applies and our limit equals 4e4x 4 x 0 4e4x (1 + 4x) 4 lim Again the numerator and denominator tend to zero and we apply l H pital s Rule; the limit equals 1 16e4x = x 0 16e4x (2 + 4x) 2 lim You Try It: Evaluate the limit limx 0 1 cos x 1 + 2 x2
OTHER ALGEBRAIC MANIPULATIONS
Sometimes a factorization helps to clarify a subtle limit: EXAMPLE 512
Evaluate the limit
x +
x 2 (x 4 + 4x 2 + 5)1/2
SOLUTION The limit as written is of the form We rewrite it as 1 (1 + 4x 2 + 5x 4 )1/2 x + x + x 2 Notice that both the numerator and denominator tend to zero, so it is now indeterminate of the form 0/0 We may thus apply l H pital s Rule The result is that the limit equals lim x 2 1 (1 + 4x 2 + 5x 4 )1/2 = lim ( 1/2)(1 + 4x 2 + 5x 4 ) 1/2 ( 8x 3 20x 5 ) x + 2x 3 lim = lim (1 + 4x 2 + 5x 4 ) 1/2 (2 + 5x 2 )
x +
Since this last limit is 2, we conclude that
x +
Indeterminate Forms
x 2 (x 4 + 4x 2 + 5)1/2 = 2
EXAMPLE 513
Evaluate
x
e x (e 3x x 4 )1/3
SOLUTION First rewrite the limit as lim e x 1 (1 x 4 e3x )1/3 = lim 1 (1 x 4 e3x )1/3 x ex
x
Notice that both the numerator and denominator tend to zero (here we use the result analogous to Example 57 that x 4 e3x 0) So our new expression is indeterminate of the form 0/0 l H pital s Rule applies and our limit equals (1/3)(1 x 4 e3x ) 2/3 ( 4x 3 e3x x 4 3e3x ) x ex lim = lim (1/3)(1 x 4 e3x ) 2/3 (4x 3 e2x + 3x 4 e2x )
x
Just as in Example 57, x 4 e3x x 3 e2x , and x 4 e2x all tend to zero We conclude that our limit equals 0 You Try It: Evaluate limx + [ x + 1 x]
Improper Integrals: A First Look
531 INTRODUCTION
The theory of the integral that we learned earlier enables us to integrate a continuous function f (x) on a closed, bounded interval [a, b] See Fig 51 However, it is frequently convenient to be able to integrate an unbounded function, or a function de ned on an unbounded interval In this section and the next we learn to do so, and we see some applications of this new technique The basic idea is that the integral of an unbounded function is the limit of integrals of bounded functions; likewise, the integral of a function on an unbounded interval is the limit of the integral on bounded intervals
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