354 Chapter 6 • Differentiable Functions
The limits included in Example 6.6.8 were chosen for their straightforward
illustrative nature. Their answers may seem predictable. In general, however,
indeterminate forms are very unpredictable. Before turning to the exercise set,
we give one more example in which the answer is somewhat surprising.
Example 6.6.9 Find lim ( x +
1
) x
X->00 X - 1Solution: Let f(x) = (x +
1
) x. Then ln f(x) = x ln (x +
1
). The function
x-1 x-1
ln x is continuous on (0, +oo). Thus, by Theorem 5.1.14 (c),
ln [ lim f(x)J = lim lnf(x) = lim x [ln(x + 1) - ln(x - 1)]
x--+oo x--+oo x--+oo
= lim ln(x + 1) - ln(x - 1).
x-+oo 1 /xBy L'H6pital's rule, this limit is equivalent to
1 1
lim ~ - ~ = lim x2(x - 1) - x2(x + 1)
x-+oo -1/x2 x-+oo -(x + l)(x - 1)
x3 - x2 - x3 - x2
lim 2
x-+oo 1 - X
-2x^2 -2
= lim --= lim ---,,- 1 --= 2.
X->00 1 - x^2 X-+00Therefore, lim ( x +
1
)x = e^2. D
X->00 X - 1EXERCISE SET 6.6- Prove Case 3 of Theorem 6.6.4.
--1 x2
- Suppose L is a (finite) real number, +oo, or -oo. Prove that
(a) lim f(x) = L {:::} lim f(-x) = L.
x-+x 0 x-+-xci
(b) lim f(x) = L {:::} lim f(-x) = L.
X---t-oo x--++oo - Use the result of Exercise 2 (a) to prove Cases 4, 5, and 6 of Theorem
6.6.4. - Prove Cases 7, 8, and 9 of Theorem 6.6.4.