Find
Both the numerator and denominator approach 0. However, x
2 — x — 12 = (x + 3)(x — 4). Hence,
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.
*-»!
CHAPTER 6
36
6.6
6.7
6.8
6.9
Find
In this case, neither the numerator nor the denominator approaches 0. In fact,
-12 and lim (x
2 -3x + 3) = 1. Hence, our limit is ^ =-12.
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
6.11
6.12
6.13
Find
As x-»0, both terms l/(x-2) and 4/(x
2 -4) "blow up" (that is, become infinitely large in magnitude). Since x2 — 4 = (x + 2)(x — 2), we can factor out
Hence, the limit reduces to
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)| suffices to choose 8 = c/2 (or any positive number smaller than e/2).
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| 5(*-3). So, we must have \5(x-3)\ of 5 would be el5.
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.
and simplify: