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Quiz 2 with Solution for Calculus I | MATH 115, Quizzes of Calculus

Material Type: Quiz; Professor: Mermin; Class: Calculus I; Subject: Mathematics; University: University of Kansas; Term: Unknown 1989;

Typology: Quizzes

Pre 2010

Uploaded on 09/17/2009

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Math 115
Jeff Merminโ€™s sections, Quiz 2, September 1
1. Compute (xโˆ’1)3.
Solution:
(xโˆ’1)3= ((xโˆ’1)(xโˆ’1)) (xโˆ’1)
= (x(xโˆ’1) โˆ’1(xโˆ’1)) (xโˆ’1)
=๎˜€x2โˆ’xโˆ’(xโˆ’1)๎˜(xโˆ’1)
= (x2โˆ’2x+ 1)(xโˆ’1)
=x2(xโˆ’1) โˆ’2x(xโˆ’1) + 1(xโˆ’1)
=x3โˆ’3x2+ 3xโˆ’1.
2. Compute 2xโˆ’2y
2.
Solution: 2xโˆ’2y
2=2(xโˆ’y)
2=2
2(xโˆ’y) = xโˆ’y.
3. Compute the following limits (or state that they do not exist), using the
method of your choice. No justification is required for correct answers.
(a) lim
xโ†’0
x3โˆ’1
2x3+x2
Solution:
lim
xโ†’0
x3โˆ’1
2x3+x2=
lim
xโ†’0x3โˆ’1
lim
xโ†’02x3+x2
=โˆ’1
0
=โˆ’โˆž since the dominant term of the denominator,
x2, is always positive near x= 0,
or, if you prefer, does not exist .
(b) lim
xโ†’1
x3โˆ’1
2x3+x2
Solution:
lim
xโ†’1
x3โˆ’1
2x3+x2=
lim
xโ†’1x3โˆ’1
lim
xโ†’12x3+x2
=0
3=0.
1
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Math 115

Jeff Merminโ€™s sections, Quiz 2, September 1

  1. Compute (x โˆ’ 1)^3.

Solution:

(x โˆ’ 1)^3 = ((x โˆ’ 1)(x โˆ’ 1)) (x โˆ’ 1) = (x(x โˆ’ 1) โˆ’ 1(x โˆ’ 1)) (x โˆ’ 1) =

x^2 โˆ’ x โˆ’ (x โˆ’ 1)

(x โˆ’ 1) = (x^2 โˆ’ 2 x + 1)(x โˆ’ 1) = x^2 (x โˆ’ 1) โˆ’ 2 x(x โˆ’ 1) + 1(x โˆ’ 1)

= x^3 โˆ’ 3 x^2 + 3x โˆ’ 1.

  1. Compute 2 xโˆ’ 2 2 y.

Solution: 2 xโˆ’ 2 2 y= 2(x 2 โˆ’ y)= 22 (x โˆ’ y) = x โˆ’ y.

  1. Compute the following limits (or state that they do not exist), using the method of your choice. No justification is required for correct answers.

(a) lim xโ†’ 0

x^3 โˆ’ 1 2 x^3 + x^2 Solution:

lim xโ†’ 0

x^3 โˆ’ 1 2 x^3 + x^2

lim xโ†’ 0 x^3 โˆ’ 1 lim xโ†’ 0

2 x^3 + x^2

= โˆ’โˆž since the dominant term of the denominator, x^2 , is always positive near x = 0, or, if you prefer, does not exist.

(b) lim xโ†’ 1

x^3 โˆ’ 1 2 x^3 + x^2 Solution:

lim xโ†’ 1

x^3 โˆ’ 1 2 x^3 + x^2

lim xโ†’ 1 x^3 โˆ’ 1 lim xโ†’ 1 2 x^3 + x^2

=

(c) lim xโ†’โˆ’โˆž

x^3 โˆ’ 1 2 x^3 + x^2 Solution:

lim xโ†’โˆ’โˆž

x^3 โˆ’ 1 2 x^3 + x^2

= lim xโ†’โˆ’โˆž

x^3 โˆ’ 1 2 x^3 + x^2

1 x^3 1 x^3

= lim xโ†’โˆ’โˆž

1 โˆ’ (^) x^13 2 + (^1) x

lim xโ†’โˆ’โˆž

x^3 lim xโ†’โˆ’โˆž

x = 12.

(d) lim xโ†’โˆž

x^3 โˆ’ 1 2 x^3 + x^2 Solution:

lim xโ†’โˆž

x^3 โˆ’ 1 2 x^3 + x^2

= lim xโ†’โˆž

x^3 โˆ’ 1 2 x^3 + x^2

1 x^3 1 x^3

= lim xโ†’โˆž

1 โˆ’ (^) x^13 2 + (^1) x

lim xโ†’โˆž

x^3 lim xโ†’โˆž

x = 12.

  1. Indicate whether the following statements are true or false. Where pos- sible, provide justification. (โ€œTrueโ€ means โ€œalways trueโ€, โ€œfalseโ€ means โ€œsometimes falseโ€.)

(a) lim xโ†’ 2

x x + 1

x โˆ’ 1

= lim xโ†’ 2

x x + 1

  • lim xโ†’ 2

x โˆ’ 1

This is true. The limit plays nicely with sums, that is, we have lim xโ†’a (f (x) + g(x)) = lim xโ†’a f (x) + lim xโ†’a g(x) whenever the two limits on the right exist, (which they do here). (b) If f is defined at a, then lim xโ†’a exists.

This is false. For example, let a = 0 and consider the split function

f (x) =

0 x โ‰ค 0 1 x > 0

If you have good thoughts about this question, please talk to me about it during office hours.