Math 1210
Quiz 1
January 10th, 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. No cell phones,
calculators, notes, or music players are allowed during the quiz.
Name:
Which 4 questions do you want to have graded?:
UID:
1. Suppose that f (x)g(x) = 1 for every x and that limx→a g(x) = 0. In the following
argument, which of the 4 steps is wrong and what is wrong with it?
1. Since f (x)g(x) = 1, taking limits on both sides we obtain
lim [f (x)g(x)] = lim 1 = 1
x→a
x→a
2. The limit of a product is the product of the limits so
h
i h
i
lim [f (x)g(x)] = lim f (x) · lim g(x)
x→a
x→a
x→a
3. Since limx→a g(x) = 0 we have
h
i h
i
lim f (x) · lim g(x) = 0
x→a
x→a
4. From all this we conclude that 0 = 1.
Solution: Step 2 is wrong. The limit of a product of two functions is the product of
the limits provided that both limits exist. The function g has limit 0 as x → a but f
doesn’t have a limit at 0. We showed this in class: since f (x)g(x) = 1 for all x, neither
1
f nor g can be zero at any point, so we can write f (x) = g(x)
and thus
1
1
=
x→a g(x)
0
lim f (x) = lim
x→a
2. Find the limit
lim
x→1−
1
1
−
x − 1 |x − 1|
Solution: Since we are computing the limit from the left, we have that x − 1 < 0 so
that |x − 1| = 1 − x and hence
1
1
2
1
1
2
lim−
−
= lim−
−
= lim−
=
x→1
x→1
x→1 x − 1
x − 1 |x − 1|
x−1 1−x
0
3. Find the limit
lim+
x→2
(x2 + 1)[[x]]
(3x − 1)2
Solution: Recall that limx→2+ [[x]] = 2 so that
lim+
x→2
(x2 + 1)[[x]]
2
(limx→2+ (x2 + 1)) · (limx→2+ [[x]])
(22 + 1) · 2
5·2
=
=
=
=
2
2
2
(3x − 1)
limx→2+ (3x − 1)
(3 · 2 − 1)
25
5
4. Find the limit
2x2 − 6xπ + 4π 2
x→π
x2 − π 2
lim
Solution: Evaluating at x = π yields 00 , so we factor the polynomials and obtain
2(x − π)(x − 2π)
2(x − 2π)
2(−π)
2x2 − 6xπ + 4π 2
=
lim
=
lim
=
= −1
x→π (x + π)(x − π)
x→π
x→π
x2 − π 2
x+π
2π
lim
5. Find the limits
lim+ [[x2 + 2x]],
x→3
lim− [[x2 + 2x]]
x→3
Solution: Evaluating the polynomial x2 + 2x at x = 3 we obtain 32 + 2 · 3 = 15. As x
approaches 3 from the right, x2 + 3x takes values which are larger than 15, so they round
down to 15 and hence limx→3+ [[x2 + 2x]] = 15. On the other hand, as we approach x = 3
from the left, x2 +3x takes values which are slightly smaller than 15, so they round down
to 14 and hence limx→3− [[x2 + 2x]] = 14
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