Review Sheet 1 Solutions
1. If f (x) = ln(x) and g(x) = x2 − 9, find expressions for the following functions and
indicate their domains.
(a) f ◦ g
(b) g ◦ f
(c) f ◦ f
(d) g ◦ g
Solution.
(a) (f ◦ g)(x) = f (g(x)) = f (x2 − 9) = ln(x2 − 9). We must have x2 − 9 > 0 for
ln(x2 − 9) to be defined, so x2 > 9. Then |x| > 3, so x > 3 or x < −3. The
domain is (−∞, −3) ∪ (3, ∞).
(b) (g ◦ f )(x) = g(f (x)) = g(ln(x)) = (ln(x))2 − 9. We must have x > 0 for ln(x) to
be defined, so the domain is (0, ∞).
(c) (f ◦ f )(x) = f (f (x)) = f (ln(x)) = ln(ln(x)). We must have x > 0 for ln(x) to be
defined and ln(x) > 0 for ln(ln(x)) to be defined. For that latter, exponentiate
both sides to see we need x > e0 = 1. So the domain is (1, ∞).
(d) (g ◦ g)(x) = g(g(x)) = g(x2 − 9) = (x2 − 9)2 − 9 = x4 − 18x2 + 72. This is a
polynomial, so its domain is the whole real line, or in other words its domain is
(−∞, ∞).
2. If f (x) = 2x + ln x, find f −1 (2).
Solution. Don’t bother trying to find f −1 (x) for general x, as this is very difficult.
Instead, note that if y = f −1 (2), then f (y) = 2. So f 1 (2) is the number that, when
input to f (x), returns 2. In other words, we need to solve
2x + ln(x) = 2.
The best way to do this is to try whole numbers near 0. That is, try x = −1, 0, 1, etc.
In this case, 1 works, since 2(1) = ln(1) = 2. So f −1 (2) = 1.
3. Find the inverse function of
f (x) =
x+1
2x + 1
Solution. Immediately write
x+1
2x + 1
and get rid of the denominator by multiplying both sides. So we’ve got y(2x + 1) =
(x + 1). Multiply through to get 2xy + y = x + 1. But now we want to isolate the x’s
on one side of the equation, since we’re solving for x. So write 2xy − x = 1 − y. Then
y=
1
we want only one copy of x in the equation, so factor out an x. Then x(2y − 1) = 1 − y.
Finally, divide by (2y − 1) to get
x=
1−y
2y − 1
4. Find the exact value of each expression.
(a) e2 ln 3
(b) log10 25 + log10 4
(c) tan(arcsin( 21 ))
(d) sin(cos−1 ( 45 ))
Solution.
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(a) e2 ln 3 = eln(3 ) = eln(9) = 9.
(b) log10 25 + log10 4 = log10 (4 · 25) = log10 100 = 2. (Since 102 = 100.)
(c) Set θ = arcsin(1/2) and draw a triangle. You know sin(theta)
= 1/2, so from the
√
triangle you’ll see that tan(arcsin(1/2)) = tan(θ) = 1/ 3.
(d) Set θ = cos−1 ( 45 ). Then cos(theta) = 4/5. Draw the triangle. The missing side
has length 3, so sin(θ) = 3/5.
5. Find the following limits.
(a) limx→−3
x2 −9
x2 +2x−3
(h−1)3 +1
h
4−v
limv→4+ |4−v|
(b) limx→0
(c)
√
(d) limx→∞ ( x2 + 4x + 1 − x)
(e) limx→0+ tan−1 (1/x)
Solution.
(a) Factor. Then the limit is equal to
lim
x→−3
(x − 3)(x + 3)
x−3
−6
3
= lim
=
=
x→−3
(x + 3)(x − 1)
x−1
−4
2
where we can divide by x + 3 since x approaches −3 but does not equal −3.
(b) Multiply it out, and you have
h3 − 3h2 + 3h − 1 + 1
h3 − 3h2 + 3h
= lim
= lim h2 − 3h + 3 = 3
x→0
x→0
x→0
h
h
lim
where we can divide by h for the same reasons as before.
(c) Since v → 4+ , v > 4 in the limit definition. Then |4 − v| = −(4 − v), since
4 − v < 0. Dividing through shows that the limit is equal to −1.
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(d) Multiply by the conjugate. We have
√
√
x2 + 4x + 1 − x2
4x + 1
x2 + 4x + 1 + x
lim ( x2 + 4x + 1−x) √ 2
= lim √ 2
= lim √ 2
x→∞
x + 4x + 1 + x x→∞ x + 4x + 1 + x x→∞ x + 4x + 1 + x
where we have used difference of√squaresqto our advantage. Now divide through
√
√
by x, recalling that y/x = y/ x2 = y/x2 provided x > 0. Then the limit is
equal to
4 + x1
lim q
x→∞
1 + x4 + x12 + 1
and since x → ∞, all terms divided by x go to zero. Then the limit is just left as
4/(1 + 1) = 2.
6. Prove that
lim x2 cos(1/x2 ) = 0
x→0
Solution. This needs the squeeze theorem. If you don’t remember what it says, look
it up. The key observation is that
−1 ≤ cos(1/x2 ) ≤ 1
because the range of cos is simply [−1, 1], regardless of what is “inside”. If we multiply
through by x2 , which is strictly positive, we have
−x2 ≤ x2 cos(1/x2 ) ≤ x2
and now the middle of this inequality is equal to our function. But because
lim −x2 = lim x2 = 0
x→0
x→0
the squeeze theorem implies that the limit of the middle quantity must also go to zero,
and this is what the problem asked.
7. Prove the statement using the precise definition of a limit.
(a) limx→2 (14 − 5x) = 4
(b) limx→0 x1/3 = 0
(c) limx→2 (x2 − 3x) = −2
(d) limx→4+
√2
x−4
=∞
Solution.
(a) We want |14 − 5x − 4| < , or |10 − 5x| < , so factoring out a 5 means we want
5|x − 2| < . We know that, for whatever δ we choose, we have |x − 2| < δ, so if
δ = /5, then 5|x − 2| < 5δ = , and we’re done. (Just write the proof up with
“Let > 0, and so on...”.)
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(b) Let’s write this up differently.
Want: |x1/3 − 0| < ,
Have: |x − 0| < δ
So now we work with what we want. We want |x1/3 | < , which is the same as
|x| < 3 . This looks the the “have” equation - in fact, if we set 3 = δ, then this
is what we have.
(c) Doing this again,
Want: |x2 − 3x − (−2)| < ,
Have: |x − 2| < δ
Again, we want to make the “want” equation look like the “have” equation. We
can rewrite the want equation as |x2 − 3x + 2| < or as |x − 2||x − 1| < . We
already have |x − 2| < δ, so we just need to do something about |x − 1|. But
really, this can’t get too big. See,
|x − 2| < δ ⇒ |x − 2| + |1| < δ + 1 ⇒ |x − 1| < δ + 1
by using the triangle inequality. So if we make sure δ < 1, then |x − 1| < 2. Then
|x − 2||x − 1| < 2δ. So if 2δ = , we’re done. The only caveat is that we need to
make sure δ < 1, so we set δ = min{/2, 1}, just to make sure.
√
(d) For the last one, there’s no δ. We want to show that |2/ x − 4| > N when
|x − 4| < δ and x > 4 (since the limit is from above) for ANY N that is given to
us. So
2 Want: √
> N, Have: |x − 4| < δ, x > 0
x − 4
Now manipulate the “want”. Remember x > 4 so the √
stuff inside the absolute
value is positive, so the “want” is really just 2/N > x − 4 or, equivalently,
(2/N )2 > x − 4 = |x − 4|, since x > 4. Then if set (2/N )2 = δ, this is just our
“have” equation, so we have what we want!
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