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GO01962-Smith-v1.cls
UAE_Math_Grade_12_Vol_1_SE_718383_ch3
July 4, 2016
Similarly, each of the following limits has the indeterminate form ∞ :
∞
( 1 ) 1 1
2
CAUTION x + 1 (x + 1) 3 x + x 3 0
2
lim = lim x ) = lim = = 0,
3
( ) x→∞ x + 5 x→∞ ( 1 x→∞ 5 1
3
We will frequently write 0 (x + 5) 1 +
0 x 3 x 3
( )
∞
or next to an expression,
∞
for instance, 3 ( 1 ) 5
0
3
lim x − 1 ( ) . x + 5 (x + 5) x 2 x + x 2
lim
x→1 x − 1 0 x→∞ x + 1 = lim ( 1 ) = lim 1 =∞
2
2
x→∞
x→∞
2
We use this shorthand to (x + 1) x 2 1 + x 2
indicate that the limit has the
indicated indeterminate form. and
This notation does not mean
0
that the value of the limit is . (2x + 3x − 5) ( 1 ) 2 + 3 − 5
2
2
You should take care to avoid 0 lim 2x + 3x − 5 = lim x 2 x x 2 = 2
writing lim f(x) = 0 or ∞ , x→∞ x + 4x − 11 x→∞ ( 1 ) = lim 4 11 1 = 2.
2
x→∞
2
x→a 0 ∞ (x + 4x − 11) 1 + −
as these are meaningless x 2 x x 2
expressions.
0
So, as with limits of the form , if a limit has the form ∞ , we must dig deeper to de-
∞
0
termine its value. Unfortunately, limits with indeterminate forms are frequently more
difficult than those just given. We struggled with the limit lim sin x (which has the
x→0 x
0
form ), ultimately resolving it only with an intricate geometric argument. In the
0 f(x)
case of lim , where lim f(x) = lim g(x) = 0, we can use linear approximations to
x→c g(x) x→c x→c
suggest a solution, as follows.
If both f and g are differentiable at x = c, then they are also continuous at x = c, so
that f(c) = lim f(x) = 0 and g(c) = lim g(x) = 0. We now have the linear approximations
x→c x→c
′
′
f(x) ≈ f(c) + f (c)(x − c) = f (c)(x − c)
′
′
and g(x) ≈ g(c) + g (c)(x − c) = g (c)(x − c),
since f(c) = 0 and g(c) = 0. As we have seen, the approximation should improve as x
approaches c, so we would expect that if the limits exist,
′
′
′
HISTORICAL lim f(x) = lim f (c)(x − c) = lim f (c) = f (c) ,
′
′
′
NOTES x→c g(x) x→c g (c)(x − c) x→c g (c) g (c)
Guillaume de l’Hôpital assuming that g (c) ≠ 0. Note that if f (x) and g (x) are continuous at x = c and g (c) ≠ 0,
′
′
′
′
(1661–1704) A French f (c) f (x)
′
′
mathematician who first then = lim . This suggests the following result.
′
′
published the result now known g (c) x→c g (x)
as l’Hôpital’s Rule. Born into
nobility, l’Hôpital was taught
calculus by the brilliant
mathematician Johann Bernoulli, THEOREM 2.1 (L’Hôpital’s Rule)
who is believed to have Suppose that f and g are differentiable on the interval (a, b), except possibly at the
discovered the rule that bears point c ∈ (a, b) and that g (x) ≠ 0on(a, b), except possibly at c. Suppose further
′
his sponsor’s name. f(x)
A competent mathematician, that lim has the indeterminate form 0 or ∞ and that
∞
x→c g(x)
0
Copyright © McGraw-Hill Education textbook. L’Hôpital was a friend lim ′ = L (or ±∞). Then, lim f(x) = lim f (x) .
l’Hôpital is best known as the
′
f (x)
author of the first calculus
x→c g (x)
and patron of many of the top
′
mathematicians of the
′
x→c g (x)
x→c g(x)
seventeenth century.
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