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MATH1901 Quizzes

Quiz 5: Limits and the limit laws
Question 1 Questions
What is the value of the limit limx1x2 x 2 x2 2x ? Exactly one option must be correct)
a)
2
b)
1
c)
1
d)
2
e)
This limit does not exist.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
limx1x2 x 2 x2 2x = 1 1 2 1 2 = 2.
Choice (e) is incorrect
What is value of the limit limx0x2 x 2 x2 2x ? Exactly one option must be correct)
a)
2
b)
1
c)
The limit does not exist
d)
e)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
We can reduce this limit to the limit limx0 1 x, which we know does not exist: limx0x2 x 2 x2 2x = limx0x2 2x + x 2 x2 2x = limx01 + x 2 x(x 2) = 1 + limx0 1 x
Choice (d) is incorrect
Choice (e) is incorrect
What is limx2x2 x 2 x2 2x ? Exactly one option must be correct)
a)
0
b)
1
c)
3 2
d)
The limit does not exist.
e)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
As in the last question, limx2x2 x 2 x2 2x = limx2x2 2x + x 2 x2 2x = limx21 + x 2 x(x 2) = 1 + limx2 1 x = 1 + 1 2.
Choice (d) is incorrect
Choice (e) is incorrect
What is limx11 + 3x 1 ? Type your answer into the box.

Correct!
limx11 + 3x 1 = 1 + 3 1 = 1.
Incorrect. Please try again.
Try substituting x = 1 into 1 + 3x 1.
What is limx0sin(3x) x ? Type your answer into the box. Exactly one option must be correct)
a)
0
b)
1 3
c)
1
d)
3
e)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Make the substitution t = 3x, so that the limit becomes
limx0sin(3x) x = limt0sin(t) t 3 = 3limt0sin(t) t = 3.
Choice (e) is incorrect
What is limx2x3 5x + 2 x3 ? Exactly one option must be correct)
a)
0
b)
1
c)
2
d)
The limit does not exist.
e)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Use the squeeze law to find limx05x2(1 cosx). Type your answer into the box.

Correct!
Since 0 1 cosx 2 we see that 0 5x2(1 cosx) 10x2. Therefore, by the squeeze law
0 limx05x2(1 cosx) lim x010x2 = 0.

Incorrect. Please try again.
Try using the inequality 1 cosx 1.
Determine limxln(x + 1) ln(x). Type your answer into the box.

Correct!
ln(x + 1) ln(x) = ln x + 1 x = ln 1 + 1 x
and limx(1 + 1 x) = ln1 = 0.

Incorrect. Please try again.
Recall that ln(a) ln(b) = ln a b.
What is lim(x,y)(0,0)x2 2y2 3x2 + y4 as (x,y) (0,0) along the xaxis? Exactly one option must be correct)
a)
0
b)
1 3
c)
1
d)

Choice (a) is incorrect
Choice (b) is correct!
Along the x–axis we have y = 0 so this limit becomes

limx0 x2 3x2 = limx01 3 = 1 3.
Choice (c) is incorrect
Choice (d) is incorrect
What is the value of the limit lim(x,y)(0,0)x2 2y2 3x2 + y4 if (x,y) approaches (0,0) along the line y = x? Exactly one option must be correct)
a)
2 3
b)
1 3
c)
0
d)

Choice (a) is incorrect
Choice (b) is correct!
Along the line y = x this limit becomes
limx0x2 2x2 3x2 + x4 = limx0 x2 3x2 + x4 = limx0 1 3 + x2 = 1 3.
Notice that Question 9 combined with Question 10 shows that the limit lim(x,y)(0,0)x2 2y2 3x2 + y4 does not exist.
Choice (c) is incorrect
Choice (d) is incorrect