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PROPERTY VI.
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If f and g are functions and lim g(x) = 0, then lim f (x) f (x) x a = x a g(x) lim g(x) lim
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The limit of a quotient is the quotient of the limits.
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CHAP. 8]
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LIMITS
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lim (2x 3 5) 2(4)3 5 123 2x 3 5 x 4 = = = lim lim (3x + 2) 3(4) + 2 14 x 4 3x + 2
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PROPERTY VII.
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f (x) =
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lim f (x)
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The limit of a square root is the square root of the limit. EXAMPLE
x2 + 5 =
lim (x 2 + 5) =
9=3
Properties IV VII have a common structure. Each tells us that, provided f and/or g has a limit as x approaches a (see Section 8.3), another, related function also has a limit as x approaches a, and this limit is as given by the indicated formula.
EXISTENCE OR NONEXISTENCE OF THE LIMIT In certain cases, a function f (x) will not approach a limit as x approaches a particular number.
EXAMPLES
(a) Figure 8-1(a) indicates that
x 0 x
does not exist. As x approaches 0, the magnitude of 1/x becomes larger and larger. (If x > 0, 1/x is positive and very large when x is close to 0. If x < 0, 1/x is negative and very small when x is close to 0.)
Fig. 8-1
(b) Figure 8-1(b) indicates that
x 0 x
LIMITS
[CHAP. 8
does not exist. When x > 0, |x| = x and |x|/x = 1; when x < 0, |x| = x and |x|/x = 1. Thus, as x approaches 0, |x|/x approaches two different values, 1 and 1, depending on whether x nears 0 through positive or through negative values. Since there is no unique limit as x approaches 0, we say that |x| x 0 x lim does not exist, (c) Let f (x) = x2 x+1 if x 1 if x > 1
Then [see Fig. 8-1(c)], lim f (x) does not exist. As x approaches 1 from the left (that is, through values of x < 1), f (x) approaches 1. But as x approaches 1 from the right (that is, through values of x > 1), f (x) approaches 2.
Notice that the existence or nonexistence of a limit for f (x) as x a does not depend on the value f (a), nor is it even required that f be de ned at a. If lim f (x) = L, then L is a number to which f (x) can be made arbitrarily close
by letting x be suf ciently close to a. The value of L or the very existence of L is determined by the behavior of f near a, not by its value at a (if such a value even exists).
Solved Problems
8.1 Find the following limits (if they exist): 1 1 (a) lim y2 (b) lim x 2 y 2 x 0 y x u2 25 u 5 u 5 (d) lim [x]
(c) lim
(a) Both y2 and 1/y have limits as y 2. So, by Property V, y2 1 y = lim y2 lim
lim 1 1 7 1 y 2 =4 =4 = y lim y 2 2 y 2
(b) Here it is necessary to proceed indirectly. The function x 2 has a limit as x 0. Hence, supposing the indicated limit to exist, Property V implies that
lim x 2 x 2
= lim
x 0 x
also exists. But that is not the case. [See example (a) in Section 8.3.] Hence, lim x2 1 x
does not exist. u2 25 (u 5)(u + 5) = lim = lim (u + 5) = 10 u 5 u 5 u 5 u 5 u 5 lim
(d) As x approaches 2 from the right (that is, with x > 2), [x] remains equal to 2 (see Fig. 7-12). However, as x approaches 2 from the left (that is, with x < 2), [x] remains equal to 1. Hence, there is no unique number that is approached by [x] as x approaches 2. Therefore, lim [x] does not exist.
CHAP. 8]
LIMITS
8.2 Find lim
f (x + h) f (x) for each of the following functions. (This limit will be important in the study of h differential calculus.) 1 (a) f (x) = 3x 1 (b) f (x) = 4x 2 x (c) f (x) = x
(a) f (x + h) = 3(x + h) 1 = 3x + 3h 1 f (x) = 3x 1 f (x + h) f (x) = (3x + 3h 1) (3x 1) = 3x + 3h 1 3x + 1 = 3h 3h f (x + h) f (x) = =3 h h f (x + h) f (x) = lim 3 = 3 h h 0
Hence,
(b) f (x + h) = 4(x + h)2 (x + h) = 4(x 2 + 2hx + h2 ) x h = 4x 2 + 8hx + 4h2 x h f (x) = 4x 2 x f (x + h) f (x) = (4x 2 + 8hx + 4h2 x h) (4x 2 x) = 4x 2 + 8hx + 4h2 x h 4x 2 + x = 8hx + 4h2 h = h(8x + 4h 1) h(8x + 4h 1) f (x + h) f (x) = = 8x + 4h 1 h h
Hence,
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