Math 165 - Quiz 2C, trig limits - solutions
Problem 1 Find the indicated limit.
2 sin(2t) + 4t cos t
t→0
3t
L = lim
Solution Split it up as in
2 sin(2t) + 4t cos t
2 sin(2t) 4 cos t
=
+
.
3t
3t
3
First,
sin(2t)
= 1.
t→0
2t
Bring this into the first half of the problem (if you like, you could multiply
both sides in the preceding equation by 2/3, or you do it in the next equation):
lim
2 sin(2t)
t→0
3t
sin(2t) 4
·
= lim
t→0
2t
3
4
4
= 1 · lim = .
t→0 3
3
L1 = lim
Second,
4 cos(t)
4
= ,
t→0
3
3
simply by substitution. Then L = L1 + L2 = 83 .
Make sure you never divide by zero in problems like this! this
is probably the fastest way to lose lots of points.
L2 = lim
Problem 2 Which choice of the constant a will make the function f (x) below
continuous everywhere?
5 − x2 for x < 2
f (x) =
ax
for x ≥ 2
Solution The function f (x) is continuous everywhere except perhaps not at
x = 2, because of the two different definitions meeting there. So we need to
ensure
lim− f (x) = f (2) = lim+ f (x).
x→2
x→2
2
For the left-hand limit, use 5 − x and substitute x = 2. For the right-hand
limit, use ax and substitute x = 2 again. So
5−4=a·2
and therefore a = 1/2 is the only choice.
Imagine that you are supposed to attach a metal rod to a curved machine
part (eg arm of a forklift). The curved part has a surface with equation
y = 5 − x2 , for x < 2, and your robot raises the metal rod with equation
y = ax (starting at a = 0, 0.1, 0.2, . . . ) until it touches the forklift arm. If
you don’t want trial and error, you need to figure out the value of a when
that happens.