CHAPTER 6
Limits
6.1 Define Km f(x) = L.
Intuitively, this means that, as x gets closer and closer to a, f(x) gets closer and closer to L. We can state this
in more precise language as follows: For any e >0, there exists 8 > 0 such that, if |*-a|<5, then
\f(x)-L\<e. Here, we assume that, for any 5 > 0 , there exists at least one x in the domain o f f ( x ) such that
\x-a\<8.
6.2 Find
The numerator and denominator both approach 0. However, u2 — 25 = (u + 5)(u — 5). Hence,
Thus,
6.3 Find
Both the numerator and denominator approach 0. However, x3 - 1 = (x - V)(x2 + x + 1). Hence,Both the numerator and denominator approach 0. However, x3 - 1 = (x - V)(x2 + x + 1). Hence,
6.4 Find
and Hence, by the quotient law for limits,
6.5 Find lim [x]. [As usual, [x] is the greatest integer s x; see Fig. 6-1.]
Fig. 6-1
As x approaches 2 from the right (that is, with * > 2), [x] remains equal to 2. However, as x approaches 2
from the left (that is, with x<2), [x] remains equal to 1. Hence, there is no unique number which is
approached by [x] as x approaches 2. Therefore, lim [x] does not exist.
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6.6 Find
Both the numerator and denominator approach 0. However, x2 — x — 12 = (x + 3)(x — 4). Hence,
6.7 Find
Both the numerator and denominator approach 0. However, division of the numerator by the denominator
reveals that x4 + 3*3 - 13*2 - 27* + 36 = (x2 + 3x- 4)(x2 - 9). Hence, lim * reveals that x4 + 3*3 - 13*2 - 27* + 36 = (x2 + 3x- 4)(x2 - 9). Hence, lim *
Hm(;c 2 -9) = l - 9 = - 8 .
*-»!
6.8 Find
In this case, neither the numerator nor the denominator approaches 0. In fact,
-12 and lim (x2 -3x + 3) = 1. Hence, our limit is ^ =-12.
6.9 Find
Both numerator and denominator approach 0. "Rationalize" the numerator by multiplying both numerator
and denominator by Vx + 3 + V3.
So, we obtain
6.10 Find
As x-»0, both terms l/(x-2) and 4/(x 2 -4) "blow up" (that is, become infinitely large in mag-
nitude). Since x2 — 4 = (x + 2)(x — 2), we can factor out and simplify:
Hence, the limit reduces to
6.11 Give an e-5 proof of the fact that
Assume e>0. We wish to find 5 >0 so that, if \x-4\<8, then \(2x-5) -3|< e. But,
(2x-5)-3 = 2x-8 = 2(x-4). Thus, we must have |2(x-4)|<e, or, equivalently, |*-4|<e/2. So, it
suffices to choose 8 = c/2 (or any positive number smaller than e/2).
6.12 In an e-S proof of the fact that lim (2'+ 5x) = 17, if we are given some e, what is the largest value of 5 that can
be used?
5 must be chosen so that, if |*-3|<S, then \(2 + 5x) -17| <e. But, (2 + 5*)-17 = 5*-15=
5(*-3). So, we must have \5(x-3)\<e, or, equivalently, |*-3|<e/5. Thus, the largest suitable value
of 5 would be el5.
6.13 Give an e-S proof of the addition property of limits: If lim f(x) = L and lim g(x) = K, then
\im(f(x) + g(x)] = L + K.
Limits
6.1 Define Km f(x) = L.
Intuitively, this means that, as x gets closer and closer to a, f(x) gets closer and closer to L. We can state this
in more precise language as follows: For any e >0, there exists 8 > 0 such that, if |*-a|<5, then
\f(x)-L\<e. Here, we assume that, for any 5 > 0 , there exists at least one x in the domain o f f ( x ) such that
\x-a\<8.
6.2 Find
The numerator and denominator both approach 0. However, u2 — 25 = (u + 5)(u — 5). Hence,
Thus,
6.3 Find
Both the numerator and denominator approach 0. However, x3 - 1 = (x - V)(x2 + x + 1). Hence,Both the numerator and denominator approach 0. However, x3 - 1 = (x - V)(x2 + x + 1). Hence,
6.4 Find
and Hence, by the quotient law for limits,
6.5 Find lim [x]. [As usual, [x] is the greatest integer s x; see Fig. 6-1.]
Fig. 6-1
As x approaches 2 from the right (that is, with * > 2), [x] remains equal to 2. However, as x approaches 2
from the left (that is, with x<2), [x] remains equal to 1. Hence, there is no unique number which is
approached by [x] as x approaches 2. Therefore, lim [x] does not exist.
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6.6 Find
Both the numerator and denominator approach 0. However, x2 — x — 12 = (x + 3)(x — 4). Hence,
6.7 Find
Both the numerator and denominator approach 0. However, division of the numerator by the denominator
reveals that x4 + 3*3 - 13*2 - 27* + 36 = (x2 + 3x- 4)(x2 - 9). Hence, lim * reveals that x4 + 3*3 - 13*2 - 27* + 36 = (x2 + 3x- 4)(x2 - 9). Hence, lim *
Hm(;c 2 -9) = l - 9 = - 8 .
*-»!
6.8 Find
In this case, neither the numerator nor the denominator approaches 0. In fact,
-12 and lim (x2 -3x + 3) = 1. Hence, our limit is ^ =-12.
6.9 Find
Both numerator and denominator approach 0. "Rationalize" the numerator by multiplying both numerator
and denominator by Vx + 3 + V3.
So, we obtain
6.10 Find
As x-»0, both terms l/(x-2) and 4/(x 2 -4) "blow up" (that is, become infinitely large in mag-
nitude). Since x2 — 4 = (x + 2)(x — 2), we can factor out and simplify:
Hence, the limit reduces to
6.11 Give an e-5 proof of the fact that
Assume e>0. We wish to find 5 >0 so that, if \x-4\<8, then \(2x-5) -3|< e. But,
(2x-5)-3 = 2x-8 = 2(x-4). Thus, we must have |2(x-4)|<e, or, equivalently, |*-4|<e/2. So, it
suffices to choose 8 = c/2 (or any positive number smaller than e/2).
6.12 In an e-S proof of the fact that lim (2'+ 5x) = 17, if we are given some e, what is the largest value of 5 that can
be used?
5 must be chosen so that, if |*-3|<S, then \(2 + 5x) -17| <e. But, (2 + 5*)-17 = 5*-15=
5(*-3). So, we must have \5(x-3)\<e, or, equivalently, |*-3|<e/5. Thus, the largest suitable value
of 5 would be el5.
6.13 Give an e-S proof of the addition property of limits: If lim f(x) = L and lim g(x) = K, then
\im(f(x) + g(x)] = L + K.
