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PROPERTY VI. in VS .NET
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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 81(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. 81 (b) Figure 81(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. 81(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. 712). 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,

