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Quiz 5: Limits and the limit laws

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
What is the value of the limit lim x2 --x---2 x→1 x2 - 2x ?
a) -2   b) -1
c) 1   d) 2
e) This limit does not exist.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
x2 --x--2 1-- 1---2 lxim→1 x2 - 2x = 1- 2 = 2.
Not correct. Choice (e) is false.
 

Question 2

 
 
What is value of the limit x2 --x-- 2 lxim→0 x2 - 2x ?
a) -2   b) 1
c) The limit does not exist   d) ∞
e) -∞

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
We can reduce this limit to the limit 1 lxim→0 x- , which we know does not exist:
limx→0 x2 --x---2 x2 - 2x = limx→0 x2 --2x+-x---2 x2 - 2x
= limx→0 1 + x - 2 x(x--2)
= 1 + limx→0 1 -- x
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 3

 
 
What is 2 lim x--2-x---2 x→2 x - 2x ?
a) 0   b) 1
c) 3 2   d) The limit does not exist.
e) ∞

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
As in the last question,
limx→2x2 - x - 2 -x2 --2x- = limx→2x2 - 2x+ x - 2 ---x2 --2x----
= limx→21 + x - 2 x(x--2)
= 1 + limx→21- x
= 1 + 1 2.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 4

 
 
What is lim √1-+-3x- 1 x→1 ? Type your answer into the box.

 

Your answer is correct
√------ √ ----- lxi→m1 1 + 3x- 1 = 1+ 3 - 1 = 1.
Not correct. You may try again.
Try substituting x = 1 into √ ------ 1+ 3x - 1.
 

Question 5

 
 
What is sin(3x) lxim→0 x ? Type your answer into the box.
a) 0   b) 1 3
c) 1   d) 3
e) ∞

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
Make the substitution t = 3x, so that the limit becomes
lim sin(3x) = lim sin(t)= 3 lim sin(t)= 3. x→0 x t→0 t3 t→0 t
Not correct. Choice (e) is false.
 

Question 6

 
 
What is lim 2x3 --5x+-2 x→ ∞ x3 ?
a) 0   b) 1
c) 2   d) The limit does not exist.
e) ∞

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 7

 
 
Use the squeeze law to find 2 lxim→0 5x(1 - cosx) . Type your answer into the box.

 

Your answer is correct
Since 0 ≤ 1 - cosx ≤ 2 we see that 0 ≤ 5x2(1 - cosx) ≤ 10x2. Therefore, by the squeeze law
0 ≤ lim 5x2(1- cosx) ≤ lim 10x2 = 0. x→0 x→0

Not correct. You may try again.
Try using the inequality -1 ≤ cosx ≤ 1.
 

Question 8

 
 
Determine ( ) xli→m∞ ln (x + 1)- ln(x) . Type your answer into the box.

 

Your answer is correct
ln(x + 1) - ln(x) = ln( x+ 1) -x--- = ln( 1 ) 1+ x-
and 1 xli→m∞ (1+ x) = ln 1 = 0 .

Not correct. You may try again.
Recall that ln(a) - ln(b) = ln a b .
 

Question 9

 
 
What is 2 2 lim x---2y-- (x,y)→ (0,0)3x2 + y4 as (x,y) → (0,0) along the x-axis?
a) 0   b) 1 3
c) 1   d) ∞

 

Not correct. Choice (a) is false.
Your answer is correct.
Along the x–axis we have y = 0 so this limit becomes

lim x2- = lim 1 = 1. x→0 3x2 x→03 3
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 10

 
 
What is the value of the limit lim x2 --2y2- (x,y)→ (0,0)3x2 + y4 if (x,y) approaches (0,0) along the line y = x?
a) 2 - - 3   b) 1 - 3
c) 0   d) ∞

 

Not correct. Choice (a) is false.
Your answer is correct.
Along the line y = x this limit becomes
x2-- 2x2 --- x2-- --- 1- 1 xli→m0 3x2 + x4 = lxim→0 3x2 + x4 = lixm→03 + x2 = - 3.
Notice that Question 9 combined with Question 10 shows that the limit 2 2 lim x---2y-- (x,y)→(0,0)3x2 + y4 does not exist.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.