1. For what values of c is f continuous on (−∞, ∞)?
(
cx2 + 1 if x ≤ 3
f (x) =
2x + c if x>3
f is continuous everywhere except 3 (dened by formula without bad
points). At 3, limx→3− f (x) = limx→3− (cx2 + 1) = c(32 ) + 1 = 9c + 1 = f (3) and
limx→3+ f (x) = limx→3+ (2x + c) = 2 · 3 + c. So both the right- and left- limit exists,
and the left one is equal to the value of the function. So f is continuous if and only if
the right limit is also equal to this value, that is 9c + 1 = 6 + c. which is equivalent
to c = 85 .
Solution:
2. Show that for some x we have f (x) = 100 if
f (x) = x3 + x sin x
f is continuous everywhere. Also f (0) = 03 + 0 sin 0 = 0 and f (10) =
1000 + 10 sin 10 ≥ 1000 − 10 = 990. Since 100 lies between f (0) and f (10), by the IVT
there is x between 0 and 10 such that f (x) = 100.
Solution:
3. Find
lim x
x→∞
Solution:
lim x
x→∞
√
√
√
x2 + a − x2 + b
We have
√
x2 + a − x 2 + b =
√
lim x
x→∞
x2 + a −
√
x2 + b
√
x2 + a +
√
√
x2 + a + x2 + b
√
x2 + b
x
√
· (x2 + a) − (x2 + b)
x→∞
x2 + a + x2 + b
a−b
p
= lim p
x→∞
1 + a/x2 + 1 + b/x2
a−b
q
= q
1 + a limx→∞ x12 + 1 + b limx→∞ x12
=
lim √
= √
a−b
a−b
√
=
.
2
1+a·0+ 1+a·0
1