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/n^{2} = π^{2}/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 ∞/∞ limitsLet

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 = 1lim (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 = LQED

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

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/

εThis 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 = 1lim (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 = LQED

Theorem: L'Hopital's RuleLet

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/y^{2} (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/y^{2})g'(x) = d/dy[g(1/y)] = g'(1/y)*d/dy[1/y] = g'(1/y)*(-1/y^{2})(9) Thus, we have:

l

im (x → +∞) [f'(x)/g'(x)] = lim (y → 0+) [f'(1/y)*(-1/y^{2})]/[g'(1/y)*(-1/y^{2})] == 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:

l

im (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