sin x < x < tan x for x in (0,π/2)

In today's blog, I will show how it is possible to use a unit circle to establish that if x is greater than 0 and less than π/2, then sin x is less than x which is less than tan x.

Theorem: if x is in (0,π/2), then sin x is less than x is less than tan x



Proof:

(1) Let O be a circle with radius = 1.

(2) Let ∠ CDO and ∠ BAO be right angles.

(3) Let x be the angle at ∠ COD

(4) We can see the following values apply to this diagram

cos x = adjacent/hypotenuse = OD/OC = OD/1 = OD

sin x = opposite/hypotenuse = CD/OC = CD/1 = CD

tan x = sin x/cos x = opposite/adjacent = AB/OA =AB/1 = AB

(5) Area of triangle OAC = (1/2)(base)(height) = (1/2)(OA)(CD) = (1/2)(1)(sin x) = (1/2)(sin x) [See Lemma 2, here for proof]

(6) Area of sector OAC = (1/2)(x)(radius)² = (1/2)(x)(1)² = (1/2)x [See Lemma 2, here for proof]

(7) Area of triangle OAB = (1/2)(base)(height) = (1/2)(OA)(AB) = (1/2)(1)(tan x) = (1/2)(tan x) [See Lemma 2, here for proof]

(8) By the diagram above, it is clear that triangle OAB is greater than sector OAC which is greater than triangle OAC.

(9) This then gives us that (1/2)(tan x) is greater than (1/2)(x) which is greater than (1/2)(sin x).

(10) Dividing all values by (1/2) gives us:

tan x is greater than x which is greater than sin x.

QED

References

• Use of Squeezing Theorem to Find Limits, analyzemath.com
  

Cauchy's Mean Value Theorem

I use Cauchy's Mean Value Theorem in my proof of L'Hopital's Rule (proof to be added later). Augustin-Louis Cauchy used this proof to as part of his effort to make calculus more rigorous. For those interested in learning more about Cauchy's role in reworking the foundations of calclulus, check out The Origins of Cauchy's Rigorous Calculus by Judith V. Grabiner.

Theorem Cauchy's Mean Value Theorem

If f(x),g(x) are continuous functions on the closed interval [a,b] and differentiable on (a,b)

Then, there exists a number c in (a,b) such that:

[f(b) - f(a)]g'(c) = [g(b)-g(a)]f'(c)

Proof:

(1) Let us define a function h(x) such that:

h(x) = [g(b) - g(a)]*[f(x) - f(a)] - [f(b) - f(a)]*[g(x) - g(a)]

(2) h(a) = h(b) = 0 since:

h(a) = [g(b) - g(a)]*[f(a) - f(a)] - [f(b) - f(a)]*[g(a) - g(a)] = [g(b) - g(a)]*0 - [f(b) - f(a)]*0 = 0

h(b) = [g(b) - g(a)]*[f(b) - f(a)] - [f(b) - f(a)]*[g(b) - g(a)] = 0

(3) h(x) is continuous in the closed interval [a,b] since:

(a) Using the Constant Law (see Lemma 1, here), we know that we can treat the values f(a),f(b),g(a),g(b), -1 as continuous functions.

(b) Using the Multiplication Law (see Lemma 3, here), we can treat -g(a), -f(a) as continuous functions.

(c) Using the Addition Law (see Lemma 2, here), we know that the following are continuous functions:

g(b) + [- g(a)] = g(b) - g(a)
f(x) + [- f(a)] = f(x) - f(a)
f(b) + [-f(a)] = f(b) - f(a)
g(x) + [-g(a)] = g(x) - g(a)

(d) Using the Multiplication Law and the Addition Law, we can see that h(x) is a continuous function

Since h(x) = [g(b)-g(a)]*[f(x) - f(a)] - [f(b) - f(a)]*[g(x) - g(a)]

(4) h(x) is also differentiable on (a,b) since:

NOTE: The detail here is parallel to the detail in step #3.

(a) We know that all constants are differentiable (see Lemma 1, here) so this means that f(a),f(b),g(a),g(b), -1 are all differentiable on (a,b)

(b) We know that the product of all differentiable functions are differentiable (see Lemma 4, here) so this means that -g(a), -f(a) are differentiable on (a,b)

(c) We also know that the addition of differentiable functions are differentiable (see Lemma 3, here) so this means that the following are all differentiable on (a,b):

g(b) + [- g(a)] = g(b) - g(a)
f(x) + [- f(a)] = f(x) - f(a)
f(b) + [-f(a)] = f(b) - f(a)
g(x) + [-g(a)] = g(x) - g(a)

(d) Finally, from the principles that we have already reviewed we know that h(x) is differentiable on (a,b) since:

h(x) = [g(b)-g(a)]*[f(x) - f(a)] - [f(b) - f(a)]*[g(x) - g(a)]

(e) We can now define h'(x)

Let u(x) = [g(b) - g(a)]*[f(x) - f(a)]

Let v(x) = [f(b) - f(a)]*[g(x) - g(a)]

u'(x) = [g(b) - g(a)][f'(x) - 0] + [0 - 0]*[f(x) - f(a)] = [g(b) - g(a)]f'(x) [See Lemma 4, here]

v'(x) = [f(b) - f(a)][g'(x) - 0] + [0 - 0]*[g(x) - g(a)] = [f(b) - f(a)]g'(x) [See Lemma 4, here]

h(x) = u(x) - v(x)

h'(x) = u'(x) - v'(x) = [g(b) - g(a)]f'(x) - [f(b) - f(a)]g'(x) [See Lemma 3, here]

(5) Using Rolle's Theorem, we know that there exists a point c such that:

h'(c) = 0 and c in (a,b)

(6) Now, combining our result for h'(x) in step #4 with step #5, we have:

h'(c) = [g(b) - g(a)]f'(c) - [f(b) - f(a)]g'(c) = 0

(7) Adding [f(b) - f(a)]g'(c) to both sides gives us:

[g(b) - g(a)]f'(c) = [f(b) - f(a)]g'(c)

QED

Corollary: Mean Value Theorem

If f(x) is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in (a,b) such that:

f(b) - f(a) = f'(c)(b-a)

Proof:

(1) Let g(x) = x

(2) Then g'(x) = 1

(3) Using Cauchy's Mean Value Theorem above, we see that there exists a value c such that:

[f(b) - f(a)]g'(c) = [g(b)-g(a)]f'(c)

(4) Since g'(c) = 1, we see that:

f(b) - f(a) = [g(b) - g(a)]f'(c)

(5) Since g(x)=x, we see that:

f(b) - f(a) = (b - a)f'(c)

QED

References

• L'Hopital's Rule, Ask Dr. Math
• C. H. Edwards, Jr. and David E. Penney, Calculus and Analytic Geometry, Prentice Hall, 1990.
  

L'Hopital's Rule

In today's blog, I present a proof for L'Hopital's Rule which is also known as L'Hospital's Rule
which states that under certain circumstances, lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x). I use it for example in the proof that ∑ 1/n² = π²/6 (Proof to be added later).

To prove this theorem, I will need to start with some lemmas that show that L'Hopital's Rule is true for specific cases.

Lemma 1: L'Hopital's Rule for 0/0 limits

Let f(x),g(x) be two functions that are differentiable in a deleted neighborhood (b,a) such that g'(x) is a nonzero, finite real number in that neighborhood, that is, when b is less than x is less than a.

If:

lim(x → a) f(x) = 0 and lim(x → a) g(x) = 0 and lim(x → a) f'(x)/g'(x) = a finite, real number

Then:

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Proof:

(1) Let f(x),g(x) be continuous functions such that f(a)=0, g(a)=0

(2) Using Cauchy's Mean Value Theorem (see Theorem, here), for any x, there exists a point c such that x is less than c which is less than a and:

[f(a) - f(x)]g'(c) = [g(a)-g(x)]f'(c)

(3) We can rerrange the equation to give us:

f'(c)/g'(c) = [f(a) - f(x)]/[g(a) - g(x)]

(4) Since f(a)=0 and g(a) = 0, this gives us:

f'(c)/g'(c) = f(x)/g(x)

(5) Let L = lim(x → a) f'(x)/g'(x) [We know that L is a finite real number from the given]

(6) Let ε be any positive real value.

(7) By definition of limits (see Definition 1, here), for ε, there exists a δ such that:

if (x - a) is between -δ and +δ, then f'(x) - L is between -ε and +ε

(8) Since c is between x and a, we can see that as x moves toward a so does c. This gives us:

lim(c → a) f'(c)/g'(c) = L since

if (c-a) is between -δ and +δ, then f'(c) - L is between -ε and +ε

(9) But from step #4, since f(x)/g(x) = f'(c)/g'(c), we can see that as x moves toward a, c likewise moves toward a and we have:

lim(x → a) f(x)/g(x) = L since:

for ε, there exists a δ such that:

if (x - a) is between δ and +δ, then (x-c) is also between -δ and +δ and since f(x)/g(x) = f'(c)/g'(c), we have that f(x)/g(x) - L is between -ε and +ε since f(c)/g'(c) - L is between -ε and +ε [See step #7]

QED

Lemma 2: L'Hopital's Rule for ∞/∞ limits

Let f(x),g(x) be two functions that are differentiable in a deleted neighborhood (b,a) such that g'(x) is a nonzero, finite real number in that neighborhood, that is, when b is less than x is less than a.

If:

lim(x → a) f(x) = ∞ and lim(x → a) g(x) = ∞ and lim(x → a) f'(x)/g'(x) = a finite, real number

Then:

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Proof:

(1) Let L = lim (x → a) f'(x)/g'(x) where L is a finite real number.

(2) Let ε be any positive value.

(3) From step #1, there exists δ such that if x -a is between -δ and δ, then f'(x) - L is between -ε and ε [Definition of limit, see here]

(4) Let b = a - δ

(5) Using Cauchy's Mean Value Theorem (see here), we know that there exists a value c such that c is in (a,b) and:

[f(a) - f(x)]g'(c) = [g(a)-g(x)]f'(c)

(6) Multiplying both sides by 1/([g'(c)][g(a) - g(x)]) gives us:

[f(a) - f(x)]/[g(a) - g(x)] = f'(c)/g'(c)

(7) Likewise, we can multiply (-1)/(-1) to both sides to get:

[f(x) - f(a)]/[g(x) - g(a)] = f'(c)/g'(c)

(8) Since c is in (a,b) and c moves toward a as x moves toward a, we have:

lim (c → a) f'(c)/g'(c) = lim (x → a) f'(x)/g'(x) = L

(9) But this means that:

lim (x → a) ([f(x) - f(a)]/[g(x) - g(a)]) = lim (c → a) f'(c)/g'(c) = L

(10) So using the definition of a limit we have:

If x - a is between -δ and +δ, then:

[f(x) - f(a)]/[g(x) - g(a)] - L is between -ε and +ε

(11) But since x is in (a,b), we know that x - a is less than a - (a - δ) = δ so we can conclude that:

[f(x) - f(a)]/[g(x) - g(a)] - L is between -ε and +ε

(12) Let h(x) = [1 - f(a)/f(x)]/[1 - g(a)/g(x)]

(13) Now,

lim (x → a) h(x)*f(x)/g(x) = lim(x → a) ([1 - f(a)/f(x)]/[1 - g(a)/g(x)]*f(x)/g(x) =

= lim (x → a) =([f(x) - f(a)]/g(x)-g(a)]) = L

(14) So, it follows that for x in (a,b), we have:

h(x)*f(x)/g(x) - L is between -ε and +ε

(15) Since lim (x → a) f(x) = ∞ and lim (x → a) g(x) = ∞, we have:

lim (x → a) [1 - f(a)/f(x)] = 1 - 0 = 1

lim (x → a) [1 - g(a)/g(x)] = 1 - 0 = 1

(16) Using the Quotient Rule for Limits (see Lemma 7, here), we have:

lim (x → a) h(x) = (lim (x → a) [1 - f(a)/f(x)])/(lim (x → a)[1 - g(a)/g(x)]) = 1/1 = 1

(17) Using the Product Rule for Limits (see Lemma 2, here), we have:

lim (x → a) h(x)*f(x)/g(x) =lim (x → a) h(x) * lim (x → a) f(x)/g(x)

(18) This means that:

lim (x → a) f(x)/g(x) = [lim (x → a) h(x)*f(x)/g(x)]/[lim (x → a) h(x)] =

= L/1 = L

QED

Lemma 3: L'Hopital's Rule for 0/0 limits where f'(x)/g'(x) has an infinite limit

Let f(x),g(x) be two functions that are differentiable in a deleted neighborhood (b,a) such that, when b is less than x is less than a.

If:

lim(x → a) f(x) = 0 and lim(x → a) g(x) = 0 and lim(x → a) f'(x)/g'(x) = +∞ or -∞

Then:

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Proof:

(1) Let f(x),g(x) be continuous functions such that f(a)=0, g(a)=0

(2) Using Cauchy's Mean Value Theorem (see Theorem, here), for any x, there exists a point c such that x is less than c which is less than a and:

[f(a) - f(x)]g'(c) = [g(a)-g(x)]f'(c)

(3) We can rerrange the equation to give us:

f'(c)/g'(c) = [f(a) - f(x)]/[g(a) - g(x)]

(4) Since f(a)=0 and g(a) = 0, this gives us:

f'(c)/g'(c) = f(x)/g(x)

(5) Let L = lim(x → a+) f'(x)/g'(x)

We can assume that L is +∞. We could make the same argument with some adjustments if L is -∞

(6) Let ε be any positive real value.

(7) For an infinite limit, for ε, there exists a δ such that:

if (x - a) is between -δ and +δ, then f'(x) is between -1/ε and +1/ε

(8) Since c is between x and a, we can see that as x moves toward a so does c. This gives us:

lim(c → a+) f'(c)/g'(c) = +∞ since

if (c-a) is between -δ and +δ, then f'(c) is between -1/ε and +1/ε

(9) But from step #4, since f(x)/g(x) = f'(c)/g'(c), we can see that as x moves toward a, c likewise moves toward a and we have:

lim(x → a+) f(x)/g(x) = +∞ since:

for ε, there exists a δ such that:

if (x - a) is between δ and +δ, then (x-c) is also between -δ and +δ and since f(x)/g(x) = f'(c)/g'(c), we have that f(x)/g(x) is between -1/ε and +1/ε since f(c)/g'(c) is between -1/ε and +1/ε [See step #7]

QED

Lemma 4: L'Hopital's Rule for ∞/∞ limits where f'(x)/g'(x) has an infinite limit

Let f(x),g(x) be two functions that are differentiable in a deleted neighborhood (b,a) such that, when b is less than x is less than a.

If:

lim(x → a) f(x) = ∞ and lim(x → a) g(x) = ∞ and lim(x → a) f'(x)/g'(x) = +∞ or -∞

Then:

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Proof:

(1) Let L = lim (x → a) f'(x)/g'(x) where L is +∞

NOTE: We can use the same argument with some modifications if L is -∞

(2) Let ε be any positive value.

(3) From step #1, there exists δ such that if x -a is between -δ and δ, then f'(x) is between -1/ε and 1/ε

This is true since we are talking about an infinite limit where 1/ε can get as close to infinity as one wishes.

(4) Let b = a - δ

(5) Using Cauchy's Mean Value Theorem (see here), we know that there exists a value c such that c is in (a,b) and:

[f(a) - f(x)]g'(c) = [g(a)-g(x)]f'(c)

(6) Multiplying both sides by 1/([g'(c)][g(a) - g(x)]) gives us:

[f(a) - f(x)]/[g(a) - g(x)] = f'(c)/g'(c)

(7) Likewise, we can multiply (-1)/(-1) to both sides to get:

[f(x) - f(a)]/[g(x) - g(a)] = f'(c)/g'(c)

(8) Since c is in (a,b) and c moves toward a as x moves toward a, we have:

lim (c → a) f'(c)/g'(c) = lim (x → a) f'(x)/g'(x) = L

(9) But this means that:

lim (x → a) ([f(x) - f(a)]/[g(x) - g(a)]) = lim (c → a) f'(c)/g'(c) = L

(10) So using the definition of a limit we have:

If x - a is between -δ and +δ, then:

[f(x) - f(a)]/[g(x) - g(a)] is between -1/ε and +1/ε

(11) But since x is in (a,b), we know that x - a is less than a - (a - δ) = δ so we can conclude that:

[f(x) - f(a)]/[g(x) - g(a)] is between -1/ε and +1/ε

(12) Let h(x) = [1 - f(a)/f(x)]/[1 - g(a)/g(x)]

(13) Now,

lim (x → a) h(x)*f(x)/g(x) = lim(x → a) ([1 - f(a)/f(x)]/[1 - g(a)/g(x)]*f(x)/g(x) =

= lim (x → a) =([f(x) - f(a)]/g(x)-g(a)]) = L

(14) So, it follows that for x in (a,b), we have:

h(x)*f(x)/g(x) is between -1/ε and +1/ε

(15) Since lim (x → a) f(x) = ∞ and lim (x → a) g(x) = ∞, we have:

lim (x → a) [1 - f(a)/f(x)] = 1 - 0 = 1

lim (x → a) [1 - g(a)/g(x)] = 1 - 0 = 1

(16) Using the Quotient Rule for Limits (see Lemma 7, here), we have:

lim (x → a) h(x) = (lim (x → a) [1 - f(a)/f(x)])/(lim (x → a)[1 - g(a)/g(x)]) = 1/1 = 1

(17) Using the Product Rule for Limits (see Lemma 2, here), we have:

lim (x → a) h(x)*f(x)/g(x) =lim (x → a) h(x) * lim (x → a) f(x)/g(x)

(18) This means that:

lim (x → a) f(x)/g(x) = [lim (x → a) h(x)*f(x)/g(x)]/[lim (x → a) h(x)] =

= L/1 = L

QED

Theorem: L'Hopital's Rule

Let f(x),g(x) be two functions that are differentiable in a deleted neighborhood a such that g'(x) is nonzero in that neighborhood.

If one of the following conditions are true:

(a) lim(x → a) f(x) = 0 and lim(x → a) g(x) = 0

(b) lim(x → a) f(x) = ∞ and lim(x → a) g(x) = ∞

Then:

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Proof:

(1) Now, we need to be able to handle the following four cases:

Case I: a is a real number/L is a real number
Case II: a is a real number/L is infinity
Case III: a is infinity/L is a real number
Case IV: a is infinty/L is infinity

(2) Case I is handled through Lemma 1 and Lemma 2.

(3) Case II is handled through Lemma 3 and Lemma 4.

(4) Assume that a is +∞

We can make the same arguments if a is -∞ with some modifications.

(5) There exists y such that 1/y = x.

(6) As x goes towards +∞, y goes toward +0.

(7) dx/dy = -1/y² (See Lemma 2, here)

(8) Using the Chain Rule (see Lemma 2, here)

f'(x) = d/dy[f(1/y)] = f'(1/y)*d/dy[1/y] = f'(1/y)*(-1/y²)

g'(x) = d/dy[g(1/y)] = g'(1/y)*d/dy[1/y] = g'(1/y)*(-1/y²)

(9) Thus, we have:

lim (x → +∞) [f'(x)/g'(x)] = lim (y → 0+) [f'(1/y)*(-1/y²)]/[g'(1/y)*(-1/y²)] =

= lim (y→ 0+) [f'(1/y)/g'(1/y)]

(10) Now, depending on f(x),g(x), f'(x)/g'(x), we can use Lemma 1, 2, 3, or 4 to establish:

lim (y → 0+) [f'(1/y)/g'(1/y)] = lim (y → 0+) f(1/y)/g(1/y)

(11) Since x=1/y, this gives us:

lim (x → +∞) [f'(x)/g'(x)] = lim(y → 0+) [f'(1/y)/g'(1/y)] = lim (y → 0+) f(1/y)/g(1/y) = lim (x → +∞) f(x)/g(x)

QED

References

• L'Hopital's Rule, Ask Dr. Math
• C. H. Edwards, Jr. and David E. Penney, Calculus and Analytic Geometry, Prentice Hall, 1990.

Products of linear factors

Using the Fundamental Theorem of Algebra, we know that it is possible to express any equation of degree n with one variable as a product of linear factors.

In other words:

xⁿ + a₁x^n-1 + ... + a_n = (x - r₁)(x - r₂)*....*(x - r_n) where r_i represent the n roots for the equation.

In today's, blog, I show that we are not limited to this form.

Lemma 1:

if (r_i ≠ 0), then:

(x - r_i) = 0 if and only if (1 - x/r_i) = 0

Proof

(1) Assume that:

(x - r_i) = 0

(2) Then dividing both sides by (-r_i) gives us:

(1 - x/r_i) = 0

(3) Assume that:

(1 - x/r_i) = 0

(4) Multiply both sides by (-r_i) so that:

(x - r_i) = 0

QED

Corollary 1.1:

if (r_i² ≠ 0), then:

(x² - r_i²) = 0 if and only if (1 - x²/r_i²) = 0

Proof:

This follows directly from Lemma 1 above if set x' = x² and r' = r_i² then we have:

(x² - r_i²) = (1 - x²/r_i²) if and only if (x' - r') = (1 - x'/r')

QED