Math 165 - Quiz 2A, trig limits - solutions
Problem 1 Find the indicated limit.
t sin(2t) + tan(5t)
t→0
3t
L = lim
Solution Split it up! First,
t sin(2t)
sin(2t)
=
,
3t
3
and the limit of this for t → 0 is zero, simply by substitution. Second,
sin(5t)
= 1.
t→0
5t
lim
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):
sin(5t)
t→0 3t cos(5t)
sin(5t)
5
= lim
·
t→0
5t
3 cos(5t)
5
5
= 1 · lim
= .
t→0 3 cos(5t)
3
L = 0 + lim
The last limit above was again done by substitution (cos(0) = 1).
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 = 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 = 4 − x2 , for x < 1, and your robot raises the metal rod with equation
y = ax + 1 (starting at a = 1, 1.1, 1.2, . . . ) 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.