Download Solutions to suggested problems

Calculus of vector valued functions
Limits and continuity - suggested problems - solutions
P1: Find limt→1 r(t) for
r(t) =<
√
t2 + t − 2
, sin(πt), 4 − t >
t−1
(t − 1)(t + 2)
t2 + t − 2
= lim
= lim(t + 2) = 1 + 2 = 3
t→1
t→1
t→1
t−1
t−1
lim sin(πt) = sin π = 0
t→1
√
√
√
lim 4 − t = 4 − 1 = 3
lim
t→1
so
lim <
t→1
√
√
t2 + t − 2
, sin(πt), 4 − t >=< 3, 0, 3 >
t−1
P2: Find limt→2− r(t) for
r(t) =< t2 + 3,
3
π
, tan >
2−t
t
lim (t2 + 3) = 22 + 3 = 7
t→2−
3
=∞
t→2− 2 − t
π
lim tan = ∞
t→2−
t
lim
3 has a vertical asymptote at t = 2,
You may want to quickly graph the last two. 2 −
t
and as t → 2− , the graph → ∞.
tan πt also has a vertical asymptote at t = 2 (or think of it as looking at the graph of
tan x when x = π2 ), and as t → 2− , the graph → ∞.
Since parts of r are unbounded, it is correct to say limt→2− r(t) does not exist. The
infinities just indicate the specific way the limit is not existing
lim < t2 + 3,
t→2−
provides that extra information.
3
π
, tan >=< 7, ∞, ∞ >
2−t
t
P3: Find limt→∞ r(t) for
r(t) =< e−t , tan−1 (t),
ln t
>
t
You’ll need to recall L’Hospital’s Rule for this one.
lim e−t = 0
t→∞
While you can’t literally plug in infinity, it’s pretty common to think e−∞ = 0, as a
shorthand for “as x → −∞, the graph of ex approaches 0”. Infinity limits come mainly
from just knowing the behavior of various functions; there’s nothing really to do, other
than graph if necessary.
π
t→∞
2
Graph it if needed; the inverse tangent function has a horizontal asymptote y = π2 .
lim tan−1 (t) =
ln t
=0
t
This is the one where you needed L’Hospital’s rule; the initial form if you “plug in” infinity
∞ = ∞ . L’Hospital says in general that if you have a function f(x) in the form ∞
is ln∞
∞
∞
g(x)
or 00
f(x)
f 0 (x)
lim
= lim 0
x→a g(x)
x→a g (x)
lim
t→∞
-differentiate the top and bottom and try again.
ln t
lim
=
t→∞ t
lim
1
t
1
1
= lim = 0
t→∞ t
So,
lim < e−t , tan−1 (t),
t→∞
t→∞
ln t
π
>=< 0, , 0 >
t
2
2
2 , sin(πt), √4 − t > continuous?
P4: On what interval(s) is r(t) =< t t+−t −
1
These are all functions which are continuous on their domains, and r is continuous on its
domain, which is the intersection of the individual domains of the component parts.
2
t−2
The domain of t +
t − 1 is {t | t 6= 1}
The domain of √
sin(πt) is (−∞, ∞)
The domain of 4 − t is {t | t ≤ 4}
Put those together, and you have that t must be less than or equal to 4, but not equal to
1. Domain of r(t) is
(−∞, 1) ∪ (1, 4]
3 , tan π > discontinuous?
P5: For what values of t is r(t) =< t2 + 3, 2 −
t
t
r is discontinuous at whatever values its component functions are discontinuous.
t2 + 3 has no discontinuities.
3
2 − t is discontinuous at t = 2.
tan x is discontinuous at odd integer multiples of π2 : x = (2k + 1) π2 . So for tan πt , let the
x be the πt , and solve for the corresponding t values:
π
π
= (2k + 1)
t
2
1
2k + 1
=
t
2
2
t =
2k + 1
tan πt is discontinuous at all values of t in the form
2 , e.g. ± 2 , ± 2 , ± 2 .
1 3 5
2k + 1
Graph this one in your spare time - it’s interesting. The vertical asymptotes get smashed
up as t → 0.
r(t) is discontinuous at {t | t =
2 , k ∈ Z}.
2k + 1