Math 1210
Quiz 2
January 17th, 2014
Answer four (4) of the five (5) questions below. Please indicate which
questions you want to have graded. Every question is worth the same. You
may use scratch paper, but you can only turn in this sheet. Please justify
your answer in the space provided: writing the solution without any
explanation is not enough. No cell phones, calculators, notes, or music
players are allowed during the quiz.
Name:
Which 4 questions do you want to have graded?:
x+1
1. Find the limits limx→2+ x2 −5x+6
and limx→2−
x+1
f (x) = x2 −5x+6 near x = 2.
UID:
x+1
.
x2 −5x+6
Sketch the graph of the function
Solution: If we evaluate the function at x = 2 we obtain 03 , so the limits will be either
x+1
. Then we
+∞ or −∞. We can factor the denominator, so that f (x) = (x−2)(x−3)
−
observe that as x → 2 , both x − 2 and x − 3 are negative, while x + 1 is positive, so
x+1
that limx→2− x2 −5x+6
= +∞. On the other hand, as x → 2+ we see that x − 2 and x + 1
x+1
are positive, while x − 3 is negative, so that limx→2+ x2 −5x+6
= −∞.
2. Find limt→0
[tan(3t)]2
.
2t
Solution:
[tan(3t)]2
1
sin(3t) sin(3t)
1
lim
=
lim lim
·
=
2
t→0
2t
2 t→0 t→0
t
cos (3t)
2
3
0
=
·1· =0
2
1
3. Find limx→∞
√2x+1
x3 +2
and limx→∞
√2x+1
x3 +2
sin
1
x4
sin(3t)
sin(3t)
3 lim
· lim
t→0
t→0 cos2 (3t)
3t
Solution: Both the numerator and the denominator of the rational function √2x+1
grow
x3 +2
to ∞ as x does. The largest exponent in the numerator is 1, whereas the largest exponent
in the denominator is 32 , so the intuition tells us that the limit will be zero. To see this we
divide both the numerator and
√ the denominator by the top degree monomial appearing
in the denominator, namely x2 , so that
2x + 1
lim √
= lim
x→∞
x3 + 2 x→∞
In order to compute limx→∞
2x + 1
lim √
sin
x→∞
x3 + 2
1
x4
√2x+1
x3 +2
√1 · (2x
x3
√
√1
x3
x3
sin
=
1
x4
+ 1)
√2
x
+
√1
x3
= lim q
x→∞
+2
1+
2
x3
=
0
=0
1
, we observe that limx→∞ sin
2x + 1
lim √
x→∞
x3 + 2
1
x4
= 0 so that
1
· lim sin
=0·0=0
x→∞
x4
4. Find the limit limx→∞
√
2x2 + 3 −
√
2x2 − 5 .
√
√
Solution: We see that as x → ∞the expression 2x2 + 3 − 2x2 − 5 goes √
to ∞ − ∞,
2x2 + 3 −
which
is
undetermined.
However
wee
see
that
multiplying
and
dividing
by
√
2x2 − 5 the indeterminacy disappears:
lim
x→∞
√
2x2 + 3 −
√
√
√2x2 + 3 + √2x2 − 5
√
√
lim
2x2 + 3 − 2x2 − 5 · √
x→∞
2x2 + 3 + 2x2 − 5
(2x2 + 3) − (2x2 − 5)
√
= lim √
x→∞
2x2 + 3 + 2x2 − 5
8
√
= lim √
2
x→∞
2x + 3 + 2x2 − 5
8
=
=0
∞
2
2x − 5 =
5. Sketch the graph of the function f (x) =
limits
[[x]]
lim−
,
t→0
x
[[x]]
x
on the interval [−1, 1] and compute the
lim+
t→0
[[x]]
x
= limx→0 x0 = 0 Note that there is no
Solution: From the right we have limx→0+ [[x]]
x
indeterminacy here: the function [[x]] is identically 0 over the interval [0, 1), so [[x]]
is
x
identically zero over [0, 1].
From the left we have limx→0−
[[x]]
x
= limx→0−
−1
x
1
= − lim− = +∞
x→0 x
| {z }
−∞
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