## Transcribed Text

1. Find the limits (a) limx→0− 2x, (b) limx→3+ x, (c) limx→3+ 5, (d) limx→−1− |x+
2|? In each case if you use a limit law, mention which one you are using.
2. Sketch by hand the graph of the function
f(x) = 1
2
x if x < −1
x if x ≥ −1
Find limx→−1− f(x). Find limx→−1+ f(x). Are these two limits the
same? What can you say about limx→−1 f(x)?
3. Use the Limit Laws to find the limit. Graph each function before you
submit the solution to check if the graph supports your conclusion. No
need to include the graph though. You could use the grapher linked from
the top right corner of this page. limx→3(x
2 − x + 2).
4. Use the Limit Laws to find the limit limx→3(x
2 − x + 2/x3
). [Note:
x
2 − x + 2/x3 6=
x
2−x+2
x3 .
5. Use the Limit Laws to find the limit limx→−3
x
2−9
x+3 .
6. Use the Limit Laws to find the limit limx→7
√
x−
√
7
x−7
7. Use the Limit Laws to find the limit limx→5
x
3−125
x−5
8. Does the limit limx→1
x
2−x−1
x−1
exist?
9. Use Limit Laws to find the limit limx→1
x
2+x−2
x−1
.
10. Graph the function f(x) =
sin π
x
if x 6= 0
0 if x = 0 where −0.2 ≤ x ≤ 0.2.
Then graph f(x) where x varies over a smaller interval, for example,
−.06 ≤ x ≤ 0.06. Determine limx→0 sin π
x
if it exists. Describe in a
paragraph the behavior of the values of f(x) near x = 0. Include graphs
in your solutions.
11. Graph the function
f(x) =
x
2
sin π
x
if x 6= 0
0 if x = 0 where − 0.2 ≤ x ≤ 0.2.
Then graph f(x) where x varies over a smaller interval, for example,
−.08 ≤ x ≤ 0.08. Describe in a paragraph the behavior of the values of
f(x) near x = 0. What are the two functions g(x) and h(x) squeezing
f(x), that is such that h(x) ≤ f(x) ≤ g(x)? What is limx→0 h(x)?
1
2
and what is limx→0 g(x)? Determine limx→0 x
2
sin π
x
if it exists. Include
graphs in your solutions.
12. Discuss on the forum but do not submit. Let f(x) be the smallest digit
that appears in the decimal expansion of x. For example, f(0.123123123123 . . .) =
0 and f(1.1111 . . .) = 1, and f(2.10101010101 . . .) = 0. Find if they
exist these limits: (a) limx→1 f(x), (b) limx→2 f(x)?
13. Discuss on the forum but do not submit. We define f on a colored
interval [0, 1).
f(x) = (
x if x is blue
1 otherwise.
(a) Determine limx→1− f(x) in the case when the entire interval [0, 1)
is blue? (b) Determine limx→1− f(x) in the case when the entire interval [0, 1) is black? (c) Does there exist limx→1− f(x) if near 1 (in
every neighborhood of 1) there are numbers colored blue and there are
numbers colored black? Hand sketch a possible graph of f.
Have a sheet of paper and a pen. This quiz is required and it is part of homework but no
need to explain your choice.
1. Based on this graph of y = x
x
,
1
bc
(A) limx→0+ x
x = 1
(B) limx→0− x
x = 1
(C) limx→0 x
x
is a number slightly less than 1.
(D) limx→0− x
x = 2.
2. limx→2−
x
3−8
x−2
(A) 12
(B) −2
(C) exists but it is equal to a number not listed here
(D) the limit does not exist since the denominator is 0 when x = 2 .
3. As x → 2
−, the values of f(x) approach −3. However, as x → 2+ the values of f(x) do not
approach any particular number. Which of the statements is then correct?
(A) limx→2− f(x) = −3 but neither limx→2+ f(x) nor limx→2 f(x) exists.
(B) limx→2+ f(x) = −3
(C) limx→2 f(x) = −3
(D) All above are incorrect.
4. Which of the following equations is correct?
(A) limh→0
h
h2 =
limh→0 h
limh→0 h2 =
0
0
2 = 0.
(B) limh→0
h
2
h =
limh→0 h
2
limh→0 h =
0
2
0 = 0.
(C) limh→0
h
2
h = limh→0 h = 0.
(D) Since the denominator goes to zero, all the above equations are not correct.
5. Let
f(x) = 1
2
x if x < −1
2x if x ≥ −1
(A) limx→−1− f(x) = −
1
2
and limx→−1+ f(x) = −2
(B) limx→−1− f(x) = −2 and limx→−1+ f(x) = −
1
2
(C) limx→−1− f(x) does not exist and limx→−1+ f(x) does not exist
(D) limx→−1− f(x) = 1 and limx→−1+ f(x) = 1
6. (A) limx→1
x
2−x+1
x+1 =
1
2
(B) limx→1
x
2−x+1
x+1 does not exist
(C) limx→1
x
2−x+1
x+1 cannot be calculated since x
2 − x + 1 does not factor
(D) none of the aboove.
7. Based on this graph of y = f(x) we conclude that
1
2
1
bc
(A) limx→0 f(x) = 1
(B) limx→0 f(x) = 0
(C) limx→0 f(x) does not exist
(D) limx→0 f(x) = 0 or limx→0 f(x) = 1.
8. Based on this graph of y = f(x) we conclude that
bc
bc
2
3
1
(A) limx→2− f(x) = 3
(B) limx→2− f(x) = 2
(C) limx→2− f(x) does not exist
(D) limx→2− f(x) = 1.

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