Unizor - Calculus - L'Hospital's Rule

Notes to a video lecture on http://www.unizor.com

L'Hopital's Rule

L'Hopital's Rule is a helpful theorem and a related technique to determine certain indeterminate form limits, like 0/0 or ∞/∞.

It should be noted that this theorem has certain limitations that are not always observed and, therefore, the method of determining the indeterminate limit based on L'Hospital's Rule might not always help.

There are, actually, a set of three theorems related to the L'Hospital's Rule, successively more powerful - basic, intermediate and advanced.

Theorem 1
(basic L'Hospital's Rule)

Given two sufficiently smooth functions, F(x) and G(x), defined on some contiguous interval with a point x=x₀ inside this interval (which means, x₀ belongs to this interval together with its immediate neighborhood).

Assume that F(x₀)=G(x₀)=0, while G(x), to be able to divide by it, is not equal to zero for x ≠ x₀. So, as x→0, both functions are infinitesimal variables and, therefore, the limit of their ratio is an indeterminate form 0/0.

Assume further that the derivative of G(x) at point x=x₀is not equal to zero:
G ^I(x₀) ≠ 0

Then
lim_x→x0 F(x)/G(x) = lim_x→x0 F ^I(x)/G ^I(x) = F ^I(x₀)/G ^I(x₀)

In short, a limit of ratio of two infinitesimal functions equals to a ratio of their derivatives at a limit point.

Analysis

Consider the ratio of two functions depicted on a graph below
F(x)=x²−2x−3 (blue) and
G(x)=−x²+7x−12 (red)
and the limit of their ratio asx→3.
Both functions approach zero asx→3, so their ratio represents an indeterminate form 0/0.
We cannot determine the limit without some clever technique.

In this particular case we can determine the limit using the following method:
F(x) = x²−2x−3 = (x−3)·(x+1)
G(x) = −x²+7x−12 = −(x−3)·(x−4)
F(x)/G(x) = [reducing by (x−3)] = (x+1)/(4−x)
And the limit of this ratio asx→3 equals to
(3+1)/(4−3) = 4

In a more general case this simple method might not work, but we will use these two functions to demonstrate a different method called L'Hospital's Rule.


We have drawn two straight lines tangentially to two given functions exactly at point x=3, where we want to find the limit of their ratio:
dark green line is tangential to F(x) and
orange line is tangential to G(x).
In the neighborhood of point of tangency x=3 function and its tangential line are very close to each other. In fact, they are so close that in the ratio of one function over another we can replace the value of a function with the value of the Y-coordinate on the tangential line for the same argument and, as we get closer to a point of tangency, the difference will be an infinitesimal that we can ignore.
And that is the main reason for L'Hospital's Rule.

Here is a justification for this.

Since
[F(x)−F(3)]/(x−3) → F ^I(3)
the difference between[F(x)−F(3)]/(x−3) and F ^I(3) is an infinitesimal variable ε, so
[F(x)−F(3)]/(x−3) = F ^I(3) + ε
from which follows
F(x)−F(3) = F ^I(3)·(x−3) + ε·(x−3)
Recall that F(3) = 0.
Therefore,
F(x) = F ^I(3)·(x−3) + ε·(x−3)
As x→3, the first member of the sum on the right hand side is an infinitesimal of the same order (this dependency is called Big-O) as x−3 because F ^I(3) is a constant. The second member is, however, an infinitesimal of a higher order (Little-o) because it's a product of two infinitesimal, ε and x−3.
Similarly,
G(x) = G ^I(3)·(x−3) + δ·(x−3)
where δ is another infinitesimal.

Using all this, the ratio of our two functions after canceling common multiplier x−3 can be written as
F(x)/G(x) = [F ^I(3) + ε]/[G ^I(3) + δ]
Since both ε and δ are infinitesimals, the limit of the expression on the right is F ^I(3)/G ^I(3).

Proof

Since F(x₀)=G(x₀)=0, we can subtract these from numerator and denominator without changing the value of a ratio.
F(x)/G(x) = [F(x)−F(x₀)] / [G(x)−G(x₀)]

Now we can divide numerator by x−x₀ to obtain an expression that resembles the definition of a derivative for function F(x):
[F(x)−F(x₀)]/(x−x₀)→F ^I(x₀) as x→x₀.
Similarly, we can divide denominator by the same x−x₀ to leave the value of a ratio unchanged and to obtain an expression in the denominator that resembles the definition of a derivative for function G(x):
[G(x)−G(x₀)]/(x−x₀)→G ^I(x₀) as x→x₀.

Now the original ratio ofF(x)/G(x) is transformed into this:
F(x)/G(x) = { [F(x)−F(x₀)]/(x−x₀) } / { [G(x)−G(x₀)]/(x−x₀) }

Both numerator and denominator of the last expression have limits as x→x₀.
They are, correspondingly, derivatives F ^I(x₀) and G ^I(x₀)with the latter not equal to zero by assumption.
Therefore, the original limit lim_x→x0 F(x)/G(x) equals to a ratio of limits, which, in turn, equals to a ratio of two derivatives at point x=x₀.
End of proof.

What if the derivative of G(x)at point x₀ is equal to zero? We cannot use this basic L'Hospital's Rule. Fortunately, we can extend this theorem to a more general case.

Theorem 2
(general L'Hospital's Rule)

Given two sufficiently smooth functions, F(x) and G(x), defined on segment [a,b].
Assume that F(x)→0 andG(x)→0 as x→+a

(which makes the limit of their ratio indeterminate as x→+a).
Assume further that G ^I(x) ≠ 0 on segment [a,b] and the limit of the ratio of the derivatives of these functions exists as x→+a.

Then
lim_x→+a F(x)/G(x) = lim_x→+a F ^I(x)/G ^I(x)
In short, a limit of ratio of two infinitesimal functions equals to a limit of ratio of their derivatives.

NOTE: This theorem does not require existence of derivatives at point x=a or the value of a derivative of G(x) not to be equal to zero at point a. Therefore, we can apply this theorem recursively, going from a ratio of functions to a ratio of their derivatives, then to a ratio of their second derivatives etc. until we get rid of indeterminate form of a limit (or give up).

Proof

Consider any point x∈(a,b) and consider our functions on interval [a,x].
We can apply Cauchy Mean Value Theorem (see the previous lecture) that states that there exists point x₀∈[a,x], where the following is true:
F ^I(x₀)/G ^I(x₀) = [F(x)-F(a)]/[G(x)-G(a)] = F(x)/G(x)

Since point x∈(a,b) can be arbitrarily chosen, let's choose x→+a.
Then point x₀ would also converge to +a since x₀∈[a,x].
That results in
lim_x→0F ^I(x₀)/G ^I(x₀) = lim_x→0F(x)/G(x)
(where x₀∈[a,x]).
Since we are dealing with sufficiently smooth functions, the last equality is equivalent to
lim_x→0F ^I(x)/G ^I(x) = lim_x→0F(x)/G(x)
End of proof.

Theorem 3
(extended L'Hospital's Rule)

Not only we can use L'Hospital's Rule to resolve indeterminate form 0/0, but also ∞/∞.
We can also extend the limit point of an argument to ±∞.

Let's prove this for a case of ∞/∞.
Assume, F(x)→∞ and G(x)→∞
as x→a.
Assume that F(x)/G(x)→L
as x→a.
We will prove that
F ^I(x)/G ^I(x)→L as well.
Recall that
[1/f(x)] ^I = −[1/f²(x)]·f ^I(x).
Therefore,
L = lim_x→aF(x)/G(x) = lim_x→a[1/G(x)]/[1/F(x)] =
(now we have 0/0 form, we can apply L'Hospital's Rule)
= lim_x→a[1/G(x)] ^I/[1/F(x)] ^I = lim_x→a [−1/G(x)]²·G ^I(x) / [−1/F(x)]²·F ^I(x) =
= lim_x→a [F(x)/G(x)]²·lim_x→aG ^I(x)/F ^I(x) = L²·lim_x→aG ^I(x)/F ^I(x)
So, we have obtained an equality
L = L²·lim_x→aG ^I(x)/F ^I(x)
from which follows that
L = lim_x→aF ^I(x)/G ^I(x)