Math 165 - Quiz 2B, trig limits - solutions
Problem 1 Find the indicated limit.
t cos(2t) + sin(5t)
t→0
3t
L = lim
Solution Split it up! First,
t cos(2t)
cos(2t)
=
,
3t
3
and the limit of this for t → 0 is 1/3, simply by substitution. Second,
lim
t→0
sin(5t)
= 1.
5t
Bring this into the second half of the problem (if you like, you could multiply
both sides in the preceding equation by 5/3, or you do it in the next equation):
1
sin(5t)
+ lim
3 t→0 3t
sin(5t) 5
1
+ lim
·
=
3 t→0 5t
3)
1 5
=
+ = 2.
3 3
L =
Make sure you never divide by zero in problems like this! this
is probably the fastest way to lose lots of points.
Problem 2 Which choice of the constant a will make the function f (x) below
continuous everywhere?
4 + x2 for x < 1
f (x) =
ax − 1 for x ≥ 1
Solution The function f (x) is continuous everywhere except perhaps not
at x = 1, because of the two different definitions meeting there. So we need
to ensure
lim− f (x) = f (1) = lim+ f (x).
x→1
x→1
2
For the left-hand limit, use 4 + x and substitute x = 1. For the right-hand
limit, use ax − 1 and substitute x = 1 again. So
4+1=a·1−1
and therefore a = 6 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 = 4 + x2 , for x < 1, and your robot raises the metal rod with equation
y = ax − 1 (starting at a = 5, 5.1, 5.2, . . . ) 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.