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Indeterminate Forms
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511 INTRODUCTION
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lim f (x) g(x) ( ) Consider the limit
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l H pital s Rule
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If limx c f (x) exists and limx c g(x) exists and is not zero then the limit ( ) is straightforward to evaluate However, as we saw in Theorem 23, when limx c g(x) = 0 then the situation is more complicated (especially when limx c f (x) = 0 as well) For example, if f (x) = sin x and g(x) = x then the limit of the quotient as x 0 exists and equals 1 However if f (x) = x and g(x) = x 2 then the limit of the quotient as x 0 does not exist In this section we learn a rule for evaluating indeterminate forms of the type ( ) when either limx c f (x) = limx c g(x) = 0 or limx c f (x) = limx c g(x) = Such limits, or forms, are considered indeterminate because the limit of the quotient might actually exist and be nite or might not exist; one cannot analyze such a form by elementary means
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CHAPTER 5 L H PITAL S RULE
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Indeterminate Forms
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Theorem 51 (l H pital s Rule) Let f (x) and g(x) be differentiable functions on (a, c) (c, b) If
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lim f (x) = lim g(x) = 0
then
f (x) f (x) = lim , x c g (x) g(x)
provided this last limit exists as a nite or in nite limit Let us learn how to use this new result EXAMPLE 51
Evaluate
x 1 x 2
ln x
+x 2
SOLUTION We rst notice that both the numerator and denominator have limit zero as x tends to 1 Thus the quotient is indeterminate at 1 and of the form 0/0 l H pital s Rule therefore applies and the limit equals (d/dx)(ln x) , x 1 (d/dx)(x 2 + x 2) lim provided this last limit exists The last limit is 1/x 1 = lim 2+x x 1 2x + 1 x 1 2x lim Therefore we see that
x 1 x 2
1 ln x = 3 +x 2
You Try It: Apply l H pital s Rule to the limit limx 2 sin( x)/(x 2 4) You Try It: Use l H pital s Rule to evaluate limh 0 (sin h/ h) and limh 0 (cos h 1/ h) These limits are important in the theory of calculus EXAMPLE 52
Evaluate the limit
x 0
x3 x sin x
CHAPTER 5 Indeterminate Forms
SOLUTION As x 0 both numerator and denominator tend to zero, so the quotient is indeterminate at 0 of the form 0/0 Thus l H pital s Rule applies Our limit equals (d/dx)x 3 , x 0 (d/dx)(x sin x) lim provided that this last limit exists It equals 3x 2 x 0 1 cos x lim This is another indeterminate form So we must again apply l H pital s Rule The result is 6x lim x 0 sin x This is again indeterminate; another application of l H pital s Rule gives us nally 6 = 6 x 0 cos x lim We conclude that the original limit equals 6 You Try It: Apply l H pital s Rule to the limit limx 0 x/[1/ ln |x|] Indeterminate Forms Involving We handle indeterminate forms involving in nity as follows: Let f (x) and g(x) be differentiable functions on (a, c) (c, b) If
lim f (x) and lim g(x)
both exist and equal + or (they may have the same sign or different signs) then f (x) f (x) = lim , x c g(x) x c g (x) lim provided this last limit exists either as a nite or in nite limit Let us look at some examples EXAMPLE 53
Evaluate the limit
x 0
lim x 3 ln |x |
SOLUTION This may be rewritten as ln |x| x 0 1/x 3 lim
Indeterminate Forms
Notice that the numerator tends to and the denominator tends to as x 0 Thus the quotient is indeterminate at 0 of the form / + So we may apply l H pital s Rule for in nite limits to see that the limit equals 1/x = lim x 3 /3 = 0 x 0 3x 4 x 0 Yet another version of l H pital s Rule, this time for unbounded intervals, is this: Let f and g be differentiable functions on an interval of the form [A, + ) If limx + f (x) = limx + g(x) = 0 or if limx + f (x) = and limx + g(x) = , then lim f (x) f (x) = lim x + g(x) x + g (x) lim provided that this last limit exists either as a nite or in nite limit The same result holds for f and g de ned on an interval of the form ( , B] and for the limit as x EXAMPLE 54