# Calculus I 2021-22 Chapter 1 (derivatives)

publicité ```Calculus 1, Chapter 1
Fall 2021
Table des mati&egrave;res
Chapter 1: Functions of one real variable: differentiability and related topics ...................................... 1
1. Inverse functions ............................................................................................................................. 1
1.1
Bijections (bijective functions, invertible functions) ............................................................. 1
1.2 Continuous bijection................................................................................................................. 3
1.3 Examples of inverse functions ................................................................................................ 3
2. Differentiability................................................................................................................................ 5
2.1. Differentiability at a point ........................................................................................................ 5
2.2 Digression: Variables vs functions ............................................................................................. 7
2.3. Applying derivatives to computation of limits ....................................................................... 10
Chapter 1: Functions of one real variable: differentiability and related
topics
1. Inverse functions
1.1
Bijections (bijective functions, invertible functions)
Definition: a bijection is a function between the elements of
two sets, where each element of one set is paired with exactly
one element of the other set, and each element of the other set
is paired with exactly one element of the first set. There are no
unpaired elements.
Definition 1.1.1 (formal):
a bijective function f: X → Y is a one-to-one (injective) and onto (surjective) mapping of a set X to a
set Y:
∀y ∈ 𝑌,
∃! 𝑥 ∈ X such that 𝑓(𝑥) = 𝑦.
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Calculus 1, Chapter 1
An injective
non surjective mapping
An injective surjective
mapping (bijection)
Fall 2021
A non-injective
surjective mapping
A non-injective nonsurjective mapping
Detecting injectivity on a graph:
Non-injective, because there are two pre-images for any strictly positive
number:
𝑓(−2) = 𝑓(2) = 4
This graph does not pass the horizontal line test
Theorem 1. 1. 1 and definition 1. 1. 2: Let f be a bijection from X to Y. Then there exists a unique
function denoted f -1, called the inverse of f, defined on X with values in Y, such that
∀𝑥 ∈ X, 𝑓 ∘ 𝑓 −1 (𝑥) = 𝑥
𝑎𝑛𝑑 ∀𝑥 ∈ Y, 𝑓
−1
∘ 𝑓(𝑥) = 𝑥.
(We can rewrite that statement in terms of functions: 𝑓 ∘ 𝑓 −1 = 𝑖𝑑𝑋 ,
identity mapping 𝑖𝑑𝐸 is defined by ∀𝑥 ∈ X, 𝑥 ↦ 𝑥)
2
𝑓 −1 ∘ 𝑓 = 𝑖𝑑𝑌 , where the
Calculus 1, Chapter 1
Fall 2021
1.2 Continuous bijection
Theorem 1. 2. 1: Any function continuous on an interval I,
with values in an interval J and strictly increasing
(respectively decreasing) on I is a bijection from I to J.
Furthermore, its inverse is a continuous function, too, and it
is strictly increasing (respectively decreasing) on J
Theorem 1. 2. 2 : In a Cartesian coordinate system, the
curves of a bijection and its inverse are symmetric to each
other with respect to the diagonal line y = x.
Here, 𝑓(𝑥) = −
1
𝑥−2
(blue line),
1
𝑔(𝑥) = 𝑓 −1 (𝑥) = 2 − 𝑥 (red line)
1.3 Examples of inverse functions
I’ll give you a couple of classical examples of inverse functions. The second example leads to a
&laquo;new&raquo; function, non-reduceable to ordinary algebraic or trigonometric functions or xponentials.
We’ll study the properties of some of such &laquo;new&raquo; functions below.
The demonstration of their existence (a consequence of theorem 1. 2. 1) is left as an exercise for
the students.
Example 1: n-th root
Definition 1. 3. 1 : Let n be a natural nonzero number. The
function defined on ℝ+ by 𝑥 ↦ 𝑥 𝑛 is a bijection from ℝ+ to
ℝ+ , which results in its being invertible. Its inverse is called
𝑛
the n-th root, denoted 𝑥 ↦ √𝑥 .
𝑛
1
Remark: we also use the notation √𝑥 = 𝑥 𝑛 .
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Calculus 1, Chapter 1
Example 2: arctan
Definition 1. 3. 2 : The function tan is is a bijection from
𝜋 𝜋
(− 2 ; 2 ) to ℝ, so that it is invertible. Its inverse function is
called arctangent, denoted arctan (in French: also arctg; in
English: also atan and tan−1 ).
Properties of arctan
arctan(0) = 0
arctan(1) =
𝜋
4
arctan(−1) = −
𝜋
4
lim arctan(𝑥) =
𝑥→+∞
𝜋
2
lim arctan(𝑥) = −
𝑥→−∞
𝜋
2
4
Fall 2021
Calculus 1, Chapter 1
Fall 2021
2. Differentiability
2.1. Differentiability at a point
Definition 2. 1. 1: Let f be a function defined in a neighborhood of a real number a. f is differentiable
at a if there exists a real numberℓ called the derivative of f at a, denoted 𝑓 ′ (𝑎) (so that 𝑓 ′ (𝑎) = ℓ)
such that:
𝑓(𝑥) − 𝑓(𝑎)
=ℓ
𝑥→𝑎
𝑥−𝑎
lim
This can also be written as lim
ℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
=ℓ
Theorem 2. 1. 1 and definition 2. 1. 2: Let f be a function defined in a neighborhood of a. Then the
following two statements are equivalent:
1. f is differentiable at a and its derivative at a is ℓ,
2. in a neighborhood of a, the following equality holds:
𝑓(𝑥) = 𝑓(𝑎) + ℓ(𝑥 − 𝑎) + (𝑥 − 𝑎)𝜀(𝑥), where lim 𝜀(𝑥) = 0
𝑥→𝑎
That equality is called a polynomial approximation of order 1 (a linear approximation) of 𝑓(𝑥) at a.
Proof (obtaining an explicit expression for 𝜀 ):
𝑓 is differentiable at 𝑎 and its derivative at 𝑎 equals ℓ:
lim
𝑥→𝑎
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
=ℓ
⇔
lim (
𝑥→𝑎
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
− ℓ)=0
Consider a function 𝜀 defined at a neighborhood of a by 𝜀(𝑥)=
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
− ℓ. Then
𝑓(𝑥)=𝑓(𝑎)+(𝑥−𝑎) ℓ +(𝑥−𝑎)𝜀(𝑥) and lim 𝜀(𝑥)=0. QCD
𝑥→𝑎
Remark: One can define in a similar way the notion of differentiability on the right and on the left of
a, using lim− 𝑓(𝑥) and lim+ 𝑓(𝑥)
𝑥→𝑎
𝑥→𝑎
Theorem 2. 1. 2: Any function differentiable at a point is continuous at that point.
Remark: The converse of that theorem is false:
Theorem 2. 1. 3: Let f be a function differentiable at a, with the derivative at a equal to ℓ. Let us
denote Cf the curve of f in an X-Y Cartesian coordinate system. Then the slope of the tangent line to
Cf drawn at the point whose X-coordinate is a, is ℓ.
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Calculus 1, Chapter 1
Remark: In a similar way, one can define left and right tangent lines
at a point where the left derivative does not equal the right
derivative. Sometimes one calls such one-side tangent lines “halftangent lines”.
Remark. If f is not differentiable at a but lim
𝑥→𝑎
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
Fall 2021
= &plusmn;∞, Cf has a vertical tangent line (or half-
tangent line) at the point whose X-coordinate is a.
Definition 2. 2. 1 : Let f be a function defined on an interval I and differentiable at all points of I.
Then we say that f is differentiable on I and we use f’ to denote the function defined on I such that at
any point x  I, f’(x) equals the derivative of f at that point.
2.2: Differentiability on an interval; derivative as a function
Definition 2. 2. 1 : Let f be a function defined on an interval I and differentiable at all points of I.
Then we say that f is differentiable on I and we use f’ to denote the function defined on I such that at
any point x  I, f’(x) equals the derivative of f at that point.
One often uses Leibniz’ notation: instead of 𝑓 ′ (𝑥) one writes
𝑑𝑓
writes 𝑑𝑥 (𝑎) or
𝑑𝑓
|
𝑑𝑥 𝑥=𝑎
𝑑𝑓
;
𝑑𝑥
. In many regards, that notation is easier to use.
In what follows it will be assumed that the formulas used to compute derivatives of basic functions
are known. If this is not the case, refresh your knowledge in any way you wish, in particular (if you
prefer) by reading the corresponding pages of the Wisconsin course available on your Moodle page.
Here, we’ll explain just how to differentiate composite functions and inverse functions.
Theorem 2. 2. 1 : The Chain Rule.
More detail: https://en.wikipedia.org/wiki/Chain_rule
Let f be a function differentiable on an interval I with values on an interval J, and let g be a function
differentiable on J. Then the composition 𝑔 ∘ 𝑓 is differentiable on I and
∀𝑥 ∈ 𝐼, (𝑔 ∘ 𝑓)′ (𝑥) = (𝑔′ ∘ 𝑓)(𝑥) &times; 𝑓 ′ (𝑥).
Why is this called “Chain Rule”? Because we can re-write this formula in Leibniz notation, where it
becomes much easier to generalize the result on compositions of more than two functions.
Let’s first prove the formula.
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Calculus 1, Chapter 1
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Proof : Let 𝑎∈I, then
(𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎)
𝑥−𝑎
=
(𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎)
𝑓(𝑥)−𝑓(𝑎)
&times;
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
=
𝑔[𝑓(𝑥)]−𝑔[𝑓(𝑎)]
𝑓(𝑥)−𝑓(𝑎)
&times; 𝑥−𝑎
𝑓(𝑥)−𝑓(𝑎)
Let 𝑋 = 𝑓(𝑥). When 𝑥 → 𝑎, 𝑋 → 𝑓(𝑎) because 𝑓 is continuous at 𝑎.
𝑔[𝑓(𝑥)]−𝑔[𝑓(𝑎)]
𝑓(𝑥)−𝑓(𝑎)
𝑥→𝑎
lim
But lim
𝑥→𝑎
𝑓(𝑥)−𝑓(𝑎)
𝑥−𝑎
Therefore, lim
𝑔(𝑋)−𝑔[𝑓(𝑎)]
𝑋−𝑓(𝑎)
𝑥→𝑎
= lim
= 𝑔’[𝑓(𝑎)] = (𝑔′ ∘ 𝑓)(𝑎) because 𝑔 is differentiable at 𝑓(𝑎).
= 𝑓′(𝑎) because 𝑓 is differentiable at 𝑎.
(𝑔∘𝑓)(𝑥)−(𝑔∘𝑓)(𝑎)
𝑥−𝑎
𝑥→𝑎
= (𝑔’ ∘ 𝑓)(𝑎) &times; 𝑓′(𝑎) QED.
In Leibniz notation, this formula can be written as
𝑑𝑔
𝑑𝑥
𝑑𝑔
= 𝑑𝑓
𝑑𝑓
𝑑𝑥
(to be compared with (𝑔 ∘ 𝑓)′ (𝑥) = (𝑔′ ∘ 𝑓)(𝑥) &times; 𝑓 ′ (𝑥) ).
which looks like as if we were dealing with fractions and then simplified by df.
When we write
𝑑𝑧
𝑑𝑥
𝑑𝑧
𝑑𝑦
= 𝑑𝑦 ⋅ 𝑑𝑥 instead of (𝑔 ∘ 𝑓)′ (𝑥) = 𝑓 ′ (𝑥)(𝑔′ ∘ 𝑓)(𝑥), it becomes easy to
generalize it to compositions of more than two functions: if 𝑧 is a function of 𝑦 which is a function of
𝑡 which is a function of 𝑥, then
𝑑𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑡
=
⋅
⋅
𝑑𝑥 𝑑𝑦 𝑑𝑡 𝑑𝑥
and so on. You can see why the rule is called the Chain Rule.
Remark. We have a peculiar situation here where the mathematical notation we
introduced is not very helpful in itself. The formula using the composition notation is
difficult to memorize. There exists a much easier way of writing the same thing, but it
uses a different notation.
d𝑦
When we re-write the chain rule in Leibniz's notation, using d𝑥 instead of y ′ (𝑥), we
consider z to be a function of the variable y which is itself a function of x (y and z are,
therefore, dependent variables), and so, z becomes a function of x as well. In that
approach, the notation z denotes a variable, not a function! z as a function of y is a
different function (mapping) than z as a function of x!
2.2 Digression: Variables vs functions
Suppose you study kinetic energy. The kinetic energy of a body of mass 𝑚 moving at a speed 𝑣 is
𝐸=
𝑚𝑣 2
.
2
We can express this same energy in terms of the momentum 𝑝 = 𝑚𝑣:
𝑝2
𝐸 = 2𝑚.
The energy 𝐸 is the same variable in the two formulas. But evidently, 𝐸 is not the same function of 𝑣
as it is a function of 𝑝 !
In Lebnitz notation, a letter denotes a variable rather than a function.
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Calculus 1, Chapter 1
Fall 2021
The Chain Rule, example 1
Consider the function
2
𝑦 = 𝑒 sin(𝑥 ) .
It can be decomposed as a composition of three functions:
𝑦 = 𝑓(𝑢) = 𝑒 𝑢
𝑢 = 𝑔(𝑣) = sin 𝑣
𝑣 = ℎ(𝑥) = 𝑥 2
𝑑𝑦
𝑑𝑢
Their derivatives are:
= 𝑓 ′ (𝑢) = 𝑒 𝑢
𝑑𝑢
= 𝑔′ (𝑣) = cos 𝑣
𝑑𝑣
𝑑𝑣
= ℎ′ (𝑥) = 2𝑥
𝑑𝑥
In Leibniz notation, the result can be written:
𝑑𝑦 𝑑𝑦
𝑑𝑢
𝑑𝑣
|
|
|
=
⋅
⋅
𝑑𝑥 𝑑𝑢 𝑢=𝑔(ℎ(𝑎)) 𝑑𝑣 𝑣=ℎ(𝑎) 𝑑𝑥 𝑥=𝑎
2
= 𝑒 sin(𝑥 ) ⋅ cos(𝑥 2 ) ⋅ 2𝑥
The Chain Rule, example 2
3
Consider the function 𝑧 = 𝑓(𝑥) = √1 − 𝑥 2 , defined on [−1, 1 ]
2
1
We can write 𝑓 = 𝑣 ∘ 𝑢 where 𝑢(𝑥) = 1 − 𝑥&sup2; and 𝑣(𝑥) = √𝑥
0
Or we can write 𝑧 = 𝑣(𝑢(𝑥)), which is the same.
-2
-1
-1
𝑢 is differentiable on [-1, 1] as a polynomial, and it has values in [0, +∞[.
-2
𝑑𝑢
-3
Moreover, ∀𝑥 ∈ [−1, 1], 𝑢′ (𝑥) = 𝑑𝑥 = −2𝑥.
𝑣 is differentiable on (0; +∞) (this is assumed to be known).
𝑑𝑣
f(x)
0
1
f'(x)
1
Moreover, ∀𝑢 ∈ (0; +∞), 𝑣 ′ (𝑢) = 𝑑𝑢 = 2 𝑢.
√
𝑑𝑓
𝑑𝑣 𝑑𝑢
𝑑𝑥
Therefore, 𝑑𝑥 = 𝑑𝑢
=
1
2√1−𝑥 2
(−2𝑥) = −
𝑥
√1−𝑥 2
If we consider that 𝑧 is the same variable whether it’s represented as a function of 𝑥 (as 𝑓(𝑥)) or a
function of 𝑢 (as 𝑣(𝑢)), we can write
𝑑𝑧
𝑑𝑥
𝑑𝑧 𝑑𝑢
𝑑𝑥
= 𝑑𝑢
which has the form stated in the Chain Rule.
A problem: 𝒗 is not differentiable on [𝟎; +∞).
Therefore, 𝑓(𝑥) is only differentiable on (-1, 1):
∀𝑥 ∈ (−1, 1), 𝑢(𝑥) &gt; 0
Therefore, 𝑓 is differentiable on (-1;1) as a composition of differentiable functions, and
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Calculus 1, Chapter 1
∀𝒙 ∈ (−𝟏, 𝟏)𝒇′ (𝒙) = 𝒖′ (𝒙) &times; 𝒗′ ∘ 𝒖(𝒙) = −𝟐𝒙 &times;
Fall 2021
𝟏
𝟐√𝟏 − 𝒙𝟐
=−
𝒙
√𝟏 − 𝒙𝟐
One cannot say anything offhand concerning the differentiability of f at 𝑥 = &plusmn;1. It has to be studied
separately.
Differentiating an inverse function
Theorem 2. 2. 2: Let f be a bijection between an interval I and an interval J, differentiable in I and
whose derivative does not become zero on J. Then f -1 is differentiable on J and its derivative is given
by
1
∀𝑥 ∈ 𝐽, (𝑓 −1 )′ (𝑥) =
.
𝑓 ′ (𝑓 −1 (𝑥))
In Leibniz notation, this can be expressed in an easy-to-memorize form:
If 𝑦 = 𝑓(𝑥) and there exists an inverse function 𝑓 −1 , then 𝑥 = 𝑓 −1 (𝑦), and
𝑑𝑥
𝑑𝑦
=
1
𝑑𝑦
𝑑𝑥
.
Example 1: The exponential function is differentiable on ℝ and is a bijection between ℝ and (0, + ∞).
Moreover, ∀𝑥 ∈ ℝ, exp′ (𝑥) = exp (𝑥) ≠ 0
The log function is the inverse of exp, therefore it is differentiable on (0, + ∞) and
∀𝑥 ∈ (0; +∞), ln′ (𝑥) =
1
1
=
exp (ln(𝑥)) 𝑥
Proof:
𝑥 = 𝑒𝑦
𝑦 = ln 𝑥
𝑑𝑥
𝑑𝑦
= 𝑒𝑦
𝑑𝑦
1
1
1
=
= 𝑦=
𝑑𝑥 𝑑𝑥 𝑒
𝑥
𝑑𝑦
𝜋 𝜋
𝜋 𝜋
Example 2: The function tan is differentiable on(− 2 ; 2 ) and is a bijection between (− 2 ; 2 ) and ℝ.
Moreover, ∀𝑥 ∈ ℝ, tan′ (𝑥) = 1 + tan&sup2;(𝑥) ≠ 0
The arctan is the inverse of function tan, therefore it is differentiable on ℝ and
∀𝑥 ∈ ℝ, arctan′ (𝑥) =
1
tan′ (arctan(𝑥))
𝑦 = arctan 𝑥
=
1
1
=
1 + tan&sup2;(arctan(𝑥)) 1 + 𝑥&sup2;
𝑥 = tan 𝑦
𝑑𝑥
𝑑𝑦
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= 1 + tan2 𝑦
Calculus 1, Chapter 1
𝑑𝑦
1
1
1
=
=
=
2
𝑑𝑥 𝑑𝑥 1 + tan 𝑦 1 + 𝑥 2
𝑑𝑦
Fall 2021
2.3. Applying derivatives to computation of limits
Theorem 2. 3. 1. L'H&ocirc;pital’s Rule
0
0
When dealing with indeterminate forms of &laquo; &raquo; or &laquo;
∞
∞
&raquo; types, it is often convenient to use
L'H&ocirc;pital’s Rule. It says that if we have a quotient of two functions: 𝑓(𝑥) and 𝑔(𝑥), and
•
lim 𝑓(𝑥) = lim 𝑔(𝑥) = 0
𝑥→𝑐
𝑥→𝑐
lim 𝑓(𝑥) = lim 𝑔(𝑥) = &plusmn;∞
or
𝑥→𝑐
𝑥→𝑐
and: (i) 𝑓 and 𝑔 are differentiable in a neighborhood of c,
(ii) 𝑔′ (𝑥) ≠ 0 in a neighborhood of c
(iii)
Then
lim
𝑓′(𝑥)
lim 𝑔′(𝑥)
𝑥→𝑐
𝑓(𝑥)
𝑥→𝑐 𝑔(𝑥)
Important!!!
exists,
= lim
𝑓′(𝑥)
𝑥→c 𝑔′(𝑥)
.
The point c can be an end of an open interval, and in particular, c can be &plusmn; ∞.
L’H&ocirc;pital’s Rule, examples
Example 1:
𝑥 2 − 2𝑥
𝑥→0 tan(𝑥)
Find lim
Set 𝑓(𝑥) = 𝑥 2 − 2𝑥 and 𝑔(𝑥) = tan(𝑥).
𝑓(0) = 0 and 𝑔(0) = 0
𝑓 is differentiable on ℝ and ∀𝑥 ∈ ℝ, 𝑓 ′ (𝑥) = 2𝑥 − 2
𝜋 𝜋
𝜋 𝜋
𝑔 is differentiable on (− 2 ; 2 ) and ∀𝑥 ∈ (− 2 ; 2 ) , 𝑔′ (𝑥) = 1 + tan&sup2;(𝑥)
Therefore, l'H&ocirc;pital’s rule yields
𝒙𝟐 − 𝟐𝒙
𝒇(𝒙)
𝒇′(𝒙) −𝟐
= 𝐥𝐢𝐦
= 𝐥𝐢𝐦
=
= −𝟐
𝒙→𝟎 𝐭𝐚𝐧(𝒙)
𝒙→𝟎 𝒈(𝒙)
𝒙→𝟎 𝒈′(𝒙)
𝟏
𝐥𝐢𝐦
Example 2:
𝑒 𝑥&sup2; −1
Find lim 𝑥 sin(𝑥)
𝑥→0
0
(&laquo; 0 &raquo;)
Set 𝑓(𝑥) = 𝑒 𝑥&sup2; − 1 and 𝑔(𝑥) = 𝑥 sin(𝑥).
𝑓(0) = 𝑔(0) = 0.
2
𝑓 ′ (𝑥) = 2𝑥𝑒 𝑥 .
𝑔′ (𝑥) = sin(𝑥) + 𝑥 cos(𝑥)
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Calculus 1, Chapter 1
𝑒 𝑥&sup2; −1
lim 𝑥 sin(𝑥)
𝑥→0
=
𝑓′(𝑥)
lim 𝑔′(𝑥)
𝑥→0
Fall 2021
2
=
2𝑥𝑒 𝑥
lim sin(𝑥)+𝑥 cos(𝑥)
𝑥→0
0
0
But this is still an indeterminate form of &laquo; &raquo; type!
Let’s use l’H&ocirc;pital on that expression:
2
𝑡(𝑥) = 2𝑥𝑒 𝑥 , 𝑠(𝑥) = sin 𝑥 + 𝑥 cos 𝑥
2
𝑡 ′ (𝑥) = 𝑓 ′′ (𝑥) = 2𝑒 𝑥 + 4𝑥 2 𝑒 𝑥&sup2;
𝑠 ′ (𝑥) = 𝑔′′ (𝑥) = cos(𝑥) cos(𝑥) − 𝑥 sin(𝑥) = 2 cos 𝑥 − 𝑥 sin 𝑥
2
2
2𝑥𝑒 𝑥
2𝑒 𝑥 +4𝑥 2 𝑒 𝑥&sup2;
2
lim sin(𝑥)+𝑥 cos(𝑥) = lim 2 cos 𝑥 −𝑥 sin 𝑥 = 2=1
𝑥→0
𝑥→0
𝑒 𝑥&sup2; −1
Result: lim 𝑥 sin(𝑥)=1
𝑥→0
You can see that in this case, to obtain a result we had to use l'H&ocirc;pital’s rule twice.
Example 3:
Find lim
1
cos (𝑥 ) − 1
𝑥→+∞
1
sin (𝑥 )
Here we are dealing with a limit at +∞, which means a one-side limit at the end of an open interval.
This does not change the calculations.
1
𝑥
′
1
𝑥
1
𝑥
1
𝑥
𝑓′(𝑥) = (cos − 1) = − ( 2 ) (− sin ) = ( 2 ) sin
1 ′
1
1
𝑥
1
𝑔′ (𝑥) = (sin 𝑥) = − (𝑥 2 ) cos 𝑥
1
sin 𝑥
𝑓′(𝑥)
=−
→
0
1 𝑥→+∞
𝑔′(𝑥)
cos 𝑥
Therefore, the limit we seek equals 0.
(It is also possible to introduce a new variable
1
𝑥
𝑦 = and calculate the limit of
cos 𝑦−1
sin 𝑦
at 𝑦 → 0+ )
L’H&ocirc;pital’s Rule, warnings
I did not put here any example where l’H&ocirc;pital’s rule won’t work. You can find several such
examples in the Wikipedia article https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule .
All of them are related to violating one of the requirements (i) – (iii) listed when formulating the
theorem.
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