Page 290 - Calculus Demystified

Chapter 2
2
s − 3s − 4 277
(f) lim = lim(s + 1) = 5.
s→4 s − 4 s→4
ln x
(g) lim = lim ln[x 1/(x−1) ]= lim ln(1 + h) 1/h = ln e = 1.
x→1 x − 1 x→1 h→0
[Here we use the non-trivial fact, explored in Chapters 5 and 6, that
lim (1 + h) 1/h = e, where e is Euler’s number.]
h→0
2
x − 9
(h) lim = lim x − 3 =−6.
x→−3 x + 3 x→−3
2. (a) lim f(x) does not exist, so f is not continuous.
x→−1
(b) lim f(x) = 1/2 and f(3) = 1/2so f is continuous at c = 3.
x→3
(c) lim f(x) = 0. If we define f(0) = 0, which is plausible from the
x→0
graph, then f is continuous at 0.
(d) lim f(x) = 0. If we define f(0) = 0, which is plausible from the
x→0
graph, then f is continuous at 0.
(e) lim f(x) = 1 and f(1) = 1so f is continuous at c = 1.
x→1
(f) lim f(x) does not exist so f is not continuous at c = 1.
x→1
(g) lim f(x) = 0 and f(0) = 0so f is continuous at c = π.
x→π
2
2
(h) lim f(x) = 2 · e and f(2) = 2 · e so f is continuous at c = 2.
x→2
3. (a) We calculate
f(2 + h) − f(2)

f (2) = lim
h→0 h
2 2
[(2 + h) + 4(2 + h)]−[2 + 4 · 2]
= lim
h→0 h
2
[4 + 4h + h + 8 + 4h]−[4 + 8]
= lim
h→0 h
2
h + 8h
= lim
h→0 h
= lim h + 8
h→0
= 8.
The derivative is therefore equal to 8.
(b) We calculate
f(1 + h) − f (h)

f (1) = lim
h→0 h