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 ∞/∞ 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/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:

lim (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:

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.

## 2 comments :

Its one of the best trick to solve the calculus problem.And in competition exams mostly the question of calculus are picked from this rule.

L Hospital's Rule

It's a good post. I really like it :). Thanks for providing good examples onL Hospital's Rule.

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