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MATH1001 Quizzes

Quiz 10: Limits, injectivity, inverse functions
Question 1 Questions
Check all of the following functions which are injective. (Zero or more options can be correct)
a)
f(x) = lnx on domain (0,)
b)
f(x) = coshx on domain [0,)
c)
f(x) = x3 + 1 on domain
d)
f(x) = sinx on domain

There is at least one mistake.
For example, choice (a) should be True.
The function is injective on this domain because its graph passes the horizontal line test, or alternatively, its derivative f(x) = 1x is positive, indicating an increasing (hence injective) function.
There is at least one mistake.
For example, choice (b) should be True.
The function is injective on this domain because its derivative f(x) = sinhx is positive for all x in (0,), indicating an increasing (hence injective) function. Note that the domain used here is not the natural domain, and has been chosen to make cosh injective.
There is at least one mistake.
For example, choice (c) should be True.
The function is injective on this domain because its graph passes the horizontal line test.
There is at least one mistake.
For example, choice (d) should be False.
The function is not injective on this domain because, for example, the distinct input values π and 2π give the same output value 0.
Correct!
  1. True The function is injective on this domain because its graph passes the horizontal line test, or alternatively, its derivative f(x) = 1x is positive, indicating an increasing (hence injective) function.
  2. True The function is injective on this domain because its derivative f(x) = sinhx is positive for all x in (0,), indicating an increasing (hence injective) function. Note that the domain used here is not the natural domain, and has been chosen to make cosh injective.
  3. True The function is injective on this domain because its graph passes the horizontal line test.
  4. False The function is not injective on this domain because, for example, the distinct input values π and 2π give the same output value 0.
If f(x) = x2 + 1 on domain (0,) and g(x) = ex on domain , find the formula for the inverse function of the composite function (f g)(x) = f(g(x)) and find its range. Exactly one option must be correct)
a)
(f g)1(x) = ln (x 1), range
b)
(f g)1(x) = ln (x) 1, range (0,)
c)
(f g)1(x) = 1 2 ln(x 1), range
d)
(f g)1(x) = e2x+1, range (0,)
e)
(f g)1(x) = 1 2 ln(x 1), range (1,)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The composite function has formula (f g)(x) = f(g(x)) = e2x + 1. It has domain and range (1,). Hence its inverse function has range and domain (1,). The formula for the inverse is found by rearranging y = e2x + 1 to make x the subject. This gives x = 1 2 ln(y 1). Hence (f g)1(x) = 1 2 ln(x 1).
Choice (d) is incorrect
Choice (e) is incorrect
Consider the function f(x) = 3 + 4x on its natural domain. Find the inverse function f1, giving its domain and range. Exactly one option must be correct)
a)
f1(x) = x2 3 4 , domain , range
b)
f1(x) = 1 3 + 4x, domain , range
c)
f1(x) = x2 3 4 , domain [0,), range [34,)
d)
f1(x) = x2 3 4 , domain , range [34,)
e)
f1(x) = x2 3 4 , domain [0,), range
f)
f1(x) = 1 3 + 4x, domain [0,), range

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The natural domain of f is [34,) and the range is [0,). These are swapped for the inverse function.
Choice (d) is incorrect
Choice (e) is incorrect
Choice (f) is incorrect
What is limx0sin(3x) x ? 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)
None of the above

Choice (a) is incorrect
Hint: divide top and bottom by x3.
Choice (b) is incorrect
Hint: divide top and bottom by x3.
Choice (c) is correct!
Choice (d) is incorrect
Hint: divide top and bottom by x3.
Choice (e) is incorrect
Hint: divide top and bottom by x3.
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.
In which of the following cases can you correctly use L’Hôpital’s rule to evaluate the limit? (Zero or more options can be correct)
a)
limxπ cosx x π
b)
limxxex
c)
limx1ln(2x) lnx
d)
limx1 lnx x 1
e)
limx0tanx x

There is at least one mistake.
For example, choice (a) should be False.
L’Hôpital’s rule cannot be used here because, although limxπx π = 0, it is not true that limxπ cosx = 0.
There is at least one mistake.
For example, choice (b) should be True.
L’Hôpital’s Rule can be used.

limxxex = lim x x ex = limx 1 ex = 0.
Note that limxx = = limxex.
There is at least one mistake.
For example, choice (c) should be False.
L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, limx1 ln(2x) = ln(2)0.
There is at least one mistake.
For example, choice (d) should be True.
Since limx1 lnx = 0 = limx1x 1 we can apply L’Hôpital’s Rule. This shows that
limx1 lnx x 1 = limx1 1 x 1 = 1.
There is at least one mistake.
For example, choice (e) should be True.
As limx0 tanx = 0 = limx0x we can apply L’Hôpital’s Rule, to find that
limx0tanx x = limx0sec2x 1 = limx01 + tan2x 1 = 0.
Correct!
  1. False L’Hôpital’s rule cannot be used here because, although limxπx π = 0, it is not true that limxπ cosx = 0.
  2. True L’Hôpital’s Rule can be used.

    limxxex = lim x x ex = limx 1 ex = 0.
    Note that limxx = = limxex.
  3. False L’Hôpital’s Rule can not be used here because although the denominator has limit 0, the numerator does not. That is, limx1 ln(2x) = ln(2)0.
  4. True Since limx1 lnx = 0 = limx1x 1 we can apply L’Hôpital’s Rule. This shows that
    limx1 lnx x 1 = limx1 1 x 1 = 1.
  5. True As limx0 tanx = 0 = limx0x we can apply L’Hôpital’s Rule, to find that
    limx0tanx x = limx0sec2x 1 = limx01 + tan2x 1 = 0.
Using L’Hôpital’s Rule find the value of limx021 + 3x 2 x . Type your answer into the box.

Correct!
Since limx021 + 3x 2 = 0 = limx0x we can apply L’Hôpital’s Rule. This gives
limx021 + 3x 2 x = limx0 3 1+3x 1 = limx0 3 1 + 3x = 3.

Incorrect. Please try again.
By L’Hôpital’s rule,

limx01 + 3x 1 x = limx0 3 21+3x 1

Using L’Hôpital’s Rule find the value of limx06ex 6x 6 x2 . Type your answer into the box.

Correct!
Notice that we can apply L’Hôpital’s Rule because limx06ex 6x 6 = 0 = lim x0x2. Therefore,

limx06ex 6x 6 x2 = limx06ex 6 2x = limx03ex 3 x Applying L’Hôpital’s rule a second time we have limx06ex 6x 6 x2 = limx03ex 1 = 3

Incorrect. Please try again.
You need to use L’Hôpital’s Rule more than once.
Find the value of the limit limx0cosx 1 x . Type your answer into the box.

Correct!
By L’Hôpital’s Rule,
limx0cosx 1 x = limx0sinx 1 = 0
Incorrect. Please try again.
Try using L’Hôpital’s Rule.