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| Selected Problems and Solutions | Tutorial №4 |
MAT194H1F
Nikita Nikolaev
Tutorial №4
Selected Problems and Solutions
MAT194H1F Calculus I for Engineering Science
Problem 1.
Show that for all x, y,
|x − y| > |x| − |y|.
Proof. We apply the triangle inequality, |a + b| 6 |a| + |b| to a = x − y and b = y, to have:
(x − y) + y 6 |x − y| + |y|, i.e. |x| 6 |x − y| + |y|.
Transferring |y| to the other side, we finally have
|x − y| > |x| − |y|.
Problem 2. Find all values of x in [0, 2π] that satisfy
2 cos x + sin 2x = 0.
Solution. Simplify:
2 cos x + sin 2x = 0,
2 cos x + 2 sin x cos x = 0,
2 cos x(1 + sin x) = 0,
which is equivalent to
cos x = 0,
π
3π
x = or
,
2
2
π
3π
x = or
,
2
2
Combining, we get that the solutions are
or
1 + sin x = 0,
or
sin x = −1,
or
x=
3π
,
.
2 2
π
1
3π
.
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
Problem 3. Find all values of x in [0, 2π] that satisfy
2 + cos 2x = 3 cos x.
Solution. Simplify:
2 + cos 2x = 3 cos x,
2 + 2 cos2 x − 1 − 3 cos x = 0,
2 cos2 x − 3 cos x + 1 = 0.
Letting y = cos x for the moment, we obtain 2y 2 − 3y + 1 = 0. Solving this quadratic
equation, we get y = 1 or y = 21 . Thus,
1
cos x = ,
2
π
5π
x = 0 or 2π,
or
or
.
3
3
Combining, we get that the solutions are 0, π3 , 5π
, 2π .
3
cos x = 1,
or
Problem 4.
What is wrong with the following equation?
x2 + x − 6
= x + 3.
x−2
(1)
Why is the following equation correct?
x2 + x − 6
= lim (x + 3).
x→2
x→2
x−2
lim
(2)
Solution. The right-hand side of (1) evaluates to 5 at x = 2, while the left-hand side of (1)
is undefined at x = 2. Therefore, equation (1) is void, unless x 6= 2, and the correct way to
write (1) would then be
x2 + x − 6
= x + 3,
if x 6= 2.
(3)
x−2
Equation (2) is correct, because on the left-hand side we have
x2 + x − 6
(x2 )(x + 3)
= lim
= lim (x + 3) = 5,
x→2
x→2
x→2
x−2
x−2
lim
and on the right-hand side,
lim (x + 3) = 5.
x→2
Thus, equation (2) is really just the equation
5 = 5,
which is obviously true for all x (because x doesn’t even appear in it).
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
Now, I know what you’re going to say: “But the left-hand side of (1) simplifies to x + 3,
which is the right-hand side. So why can’t we do that and Bob’s your uncle?” You see, one
way to make sense of the notion of an equation is to realise that when we say “one expression
is equal to another expression”, we mean that the values of one expression coincide with the
values of the other expression. The problem with this description is, of course, that it
couldn’t be more vague, and mathematics won’t stand for it, you see.
We can mend the situation by thinking of these expressions as functions, in this case
functions of x. In this light, we see that an equation is really a statement about the equality of
functions. Now, by definition, two functions are equal if their values are the same everywhere
on their domain of definition. This last bit is a very crucial part, often forgotten about by
students.
In our example, let view the left- and the right-hand sides of equation (1) are functions
L(x) and R(x), respectively:
L(x) :=
x2 + x − 6
,
x−2
R(x) := x + 3,
so (1) is just L(x) = R(x). Now, obviously, away from x 6= 2,
L(x) =
x2 + x − 6
= x + 3 = R(x),
x−2
so these functions agree (i.e. have the same values) on (−∞, 2) ∪ (2, ∞). The problematic
point is, of course, x = 2, so the domain of L(x) is (−∞, 2) ∪ (2, ∞), yet the domain of R(x)
is R. Therefore, we conclude that the two functions L(x) and R(x) are not the same, simply
because their domains are different.
e
However, let R(x)
be a new function, defined by restricting the domain of R(x) to
(−∞, 2) ∪ (2, ∞). Now, this time around, this function is equal to L(x). Indeed, funce
tions L(x) and R(x)
have the same domain, namely (−∞, 2) ∪ (2, ∞), and their values on
e
e
this domain coincide (i.e. for any x ∈ (−∞, 2) ∪ (2, ∞), L(x) = R(x)),
so L(x) equals R(x)
by definition. This is way (3) is now correct.
Then you’re going to ask: “But, then, how come we were able to do the simplification
in (2)?” You see, the process of taking a limit of a function examines the behaviour of the
function near a point, but entirely ignoring what actually happens at the point. In equation
(2), we are not at all concerned about the behaviour of L(x) and R(x) at the point x = 2,
but instead we are interested in what happens near the point x = 2 and we examine this
behaviour by testing all kinds of values of x that are close but not equal to x = 2. Therefore,
cancelling out the (x − 2) monomial on the left-hand side is well-justified.
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
Problem 5.
Evaluate the following limit, if it exists.
t2 − 9
.
t→−3 2t2 + 7t + 3
lim
Solution. Simply, start simplifying to get:
(t − 3)
t−3
−6
6
t2 − 9
(t
+3)
= lim
= lim
=
= .
lim 2
t→−3 (2t + 1)
t→−3 2t + 7t + 3
(t
+ 3) t→−3 2t + 1
−5
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Problem 6.
Evaluate the following limit, if it exists.
lim
t→0
1
1
− 2
t t +t
.
Solution. Simply, start simplifying to get:
1
t2 + t − t
t2
1
1
lim
− 2
= lim
=
lim
= lim
= 1.
2
2
t→0
t→0 t(t + t)
t→0 t (t + 1)
t→0 t + 1
t t +t
Problem 7.
If 4x − 9 6 f (x) 6 x2 − 4x + 7 for x > 0, find lim f (x).
x→4
Solution. We have lim (4x − 9) = 16 − 9 = 7, while lim (x2 − 4x + 7) = 7, so by the squeeze
x→4
x→4
theorem,
lim f (x) = 7.
x→4
Problem 8.
Evaluate the following limit, if it exists.
lim
t→0+
1
1
−
x |x|
.
Solution. Recall that “x → 0+ ” means “approach 0 from ‘above”’, and “above” here just
means “positive numbers”. Since |x| = x for x > 0, we have
1
1
1 1
lim
−
= lim+
−
= 0.
t→0+
t→0
x |x|
x x
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| Selected Problems and Solutions | Tutorial №4 |
MAT194H1F
Nikita Nikolaev
Problem 9. Prove or disprove the following statement: the following function is continuous:
(
x2 ,
if x < 1,
f (x) = √
x, if x > 1.
A quick moment of thought and we can imagine what this function roughly looks like:
Thus, it seems that a reasonable guess is that f (x) is continuous.
√
Proof. Obviously, for x < 1 and x > 1 the function is, respectively, simply x2 and x
and hence
√ continuous. Thus, it suffices to show that f (x) is continuous at x = 1. Firstly,
f (x) = 1 = 1, so it remains to show that lim f (x) = 1.
x→1
The function f (x) is piece-wise defined, so it is convenient to split the limit into the
“x → 1− ” and “x → 1+ ” parts, and show that they are equal. Now, for x < 1, f (x) = x2 , so
lim− f (x) = lim− x2 = 1,
x→1
while f (x) =
√
x→1
x for x > 1, so
lim+ f (x) = lim+
x→1
√
x = 1.
x→1
The upper and the lower limits agree, hence lim f (x) = 1. Thus, f (x) is indeed continuous.
x→1
Problem 10.
The gravitational force exerted by the planet Earth on a unit mass at a distance r from the
centre of the planet is

GM r


, if r < R,

 R3
F (r) =



 GM , if r > R,
r2
where M is the Earth’s mass, R is its radius, and G is the gravitational constant. Is F a
continuous function of r?
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
Clearly F (r) is continuous away r = R. The function F (r) is a physical quantity, and
the radius r = R corresponds to the surface of the Earth, i.e. where we live. If F (r) were
discontinuous at r = R, you can imagine that something atrociously horrible would be going
on around the surface of the earth. For example, it only takes a tiny jump to get viciously
thrown up in the air; certainly, dancing would be very difficult indeed! Thankfully, none
of this seems to happen in reality, so we get the impression that a reasonable enough guess
would be that the function F (r) should be continuous.
Proof. Note first of all that r is a physical quantity, whose meaning is the radial distance
from the centre of the Earth; so r > 0.
As we have already mentioned, if F (r) is certainly continuous in the regions r < R and
r > R, so we need only check the case r = R. Firstly,
F (R) =
GM
.
R2
Splitting the limit as we did in the previous problem, we get
lim− F (r) =
r→R
GM
GM
GM
lim− r = 3 R = 2 ,
3
R r→R
R
R
and
lim+ F (r) = GM lim+
r→R
r→R
GM
1
= 2 .
2
r
R
3
Notice that we factored out GM/R in the first limit and GM in the second limit, because
none of them depends on r. Thus, the two limits agree, so
lim F (r) =
r→R
GM
= F (R),
R2
where F (r) is indeed continuous.
Problem 11.
If f (x) = x2 + 10 sin x, show that there is a number c such that f (c) = 1000.
Proof. Firstly, notice that f is continuous as a sum of continuous functions. Next f (0) = 0
and
f (100) = 1002 + 10 · sin(100) > 1002 − 10 = 10000 − 10 > 1000,
where in our estimation we’ve used the fact that −1 6 sin x 6 1 for any x. So, by the Mean
Value Theorem, we obtain the desired result.
Problem 12.
For what values of x is the following function continuous?
(
0, if x is rational,
f (x) =
1, if x is irrational.
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
We claim that f is everywhere discontinuous (i.e. f is discontinuous at every point).
First, here’s a delightful fact.
Fact: Both rational and irrational points are dense in the real numbers R, i.e. for any
point x ∈ R, we can always find a rational point r that is arbitrarily close to x, and also
we can always find an irrational point ρ that is arbitrarily close to x. In the language of
mathematics, this means that for any x ∈ R and any number δ > 0, there exists a rational
point r, such that |x − r| < δ; there also exists an irrational point ρ, such that |x − ρ| < δ.
Using quantifiers, we can write:
∀x ∈ R, ∀δ > 0, ∃r ∈ Q, such that |x − r| < δ,
and ∃ρ ∈ (R \ Q), such that |x − ρ| < δ.
(The symbol Q denotes the set of all rational numbers; every real number that is not rational
is (essentially by definition) an irrational number, so the symbol R \ Q denotes the set of all
irrational numbers.)
Claim. f is everywhere discontinuous.
Proof. Choose any x ∈ R and assume that x is a rational number, x ∈ Q. Suppose towards
a contradiction that f is continuous at x. Then for any given ε > 0, we can always find some
δ > 0, such that, for any y that is no further than δ away from x (i.e. for any y satisfying
|x − y| < δ), we must have |f (x) − f (y)| < ε. In particular, if we choose ε = 21 , then there is
some δ > 0, such that
1
|f (x) − f (y)| < ,
2
for all y satisfying |x − y| < δ. However, by the density of irrational numbers in R (see fact
above), there exists some irrational number ρ, such that |x − ρ| < δ; but then
|f (x) − f (ρ)| = |0 − 1| = 1,
which is certainly not less than 12 . Contradiction. f is discontinuous at x. As x was chosen
arbitrarily (but rational), f is discontinuous at every rational point.
If x is irrational, then, you see, the arguments above translate nearly word-for-word,
showing that f is indeed discontinuous at every irrational point, too. Thus, f is discontinuous
everywhere.
The above proof had many explanations to aid your understanding of the arguments
involved. To help you also understand how solutions and proofs are written down properly,
with an adequate amount of justification, I include a succinct version of the above proof. It
is this kind of style that a student, not unlike yourself, should use exclusively when writing
down a solution or a proof to a homework, quiz, test or exam exercise.
Solution. Claim: f is discontinuous everywhere.
Proof. Pick any x ∈ R and assume first that x is rational. Suppose f is continuous at x.
Then for ε = 12 , there exists some δ > 0, such that, for all y satisfying |x − y| < δ,
1
|f (x) − f (y)| < .
2
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MAT194H1F
| Selected Problems and Solutions | Tutorial №4 |
Nikita Nikolaev
Since irrational numbers are dense in R, there exists some irrational ρ, such that |x − ρ| < δ.
But then
1
|f (x) − f (ρ)| = |0 − 1| = 1 > .
2
Contradiction. Thus, f is discontinuous at x.
Similarly, if x is irrational, then f is again discontinuous at x. Since x was chosen
arbitrarily, f is discontinuous at all x ∈ R.
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