L'Hopital's Rule

What is the proof of L'Hopital's Rule and what's the intuitive idea\logic behind it? Can somebody please tell me?

Note by Shubham Srivastava
5 years, 3 months ago

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To prove l'Hôpital's rule, we could use Cauchy's mean value theorem, which in turn we can prove using Rolle's theorem. I will give you proofs of all the three theorems, hopefully Rolle's and Cauchy's are a bit more intuitive and will help make it more clear how we can use them to prove l'Hôpital's. So let's start with Rolle's theorem.

Rolle's theorem. Considering a real, continuous and differentiable function \(f(x)\) on the interval \([a, b]\), if \(f(a) = f(b)\), then there exists a point \(c, \ a < c < b\), such that \(f'(c) = 0\).

Proof. Let \(M\) and \(m\) be the local maximum and minimum of \(f(x)\) on the interval \([a, b]\) (the function has to have them since it is continuous). If the maximum and minimum are both on the endpoints, i.e. \(M = m = f(a) = f(b)\), then \(f(x) = M\) for \(x \in [a, b]\), hence \(f'(x) = 0\) in the entire interval, so we can take any point in \((a, b)\) as our \(c\).

Now assume that there exists a point \(c \in (a, b)\) such that \(f(c) = M\). Since the maximum is in \(c\), for any \(h > 0\) such that \(c + h < b\), it holds that \(f(c + h) < M = f(c)\). Hence the right derivative in \(c\) must be lesser than or equal to zero. We can see this from the definition of the right derivative in \(c\):

\(f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \leq 0\)

Similarly, for any \(h < 0\), such that \(c + h > a\), it holds that \(f(c+h) < f(c)\), so the left derivative must be greater than or equal to zero:

\(f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \geq 0\)

And since the function \(f(x)\) is differentiable in \(c\), it must hold that \(f'(c^+) = f'(c^-) = 0\). We can prove it similarly for \(f(c) = m\). QED.

Now that we have that, it is easy to prove Cauchy's theorem:

Cauchy's theorem. Considering two real functions \(f(x)\) and \(g(x)\) that are continuous and differentiable on the interval \([a, b]\), there exists a \(c \in (a, b)\), such that \((f(b) - f(a)) \ g'(c) = (g(b)-g(a)) \ f'(c)\).

Proof. Let \(h(x) = f(x) - \lambda g(x)\), where \(\lambda \in \mathbb{R}\). Clearly, \(h(x)\) is both continuous and differentiable on \([a, b]\). Consider a value of \(\lambda\) such that \(h(a) =h(b)\):

\(f(a) - \lambda g(a) = f(b) - \lambda g(b)\)

\(\lambda (g(a) - g(b)) = f(a) - f(b)\)

\(\lambda = \frac{f(a) - f(b)}{g(a) - g(b)}\)

Since for this \(\lambda, h(a) = h(b)\), the function \(h(x)\) now satisfies Rolle's theorem. Hence there exists \(c \in (a, b)\) such that \(h'(c) = 0\).

\(h'(c) = f'(c) - \lambda g'(c)\)

\(\lambda = \frac{f'(c)}{g'(c)} = \frac{f(a) - f(b)}{g(a) - g(b)} \implies (g(a) - g(b)) \ f'(c) = (f(a) - f(b)) \ g'(c)\),

which is what we wanted to prove. QED.

Finally, let's move on to l'Hôpital's rule:

l'Hôpital's rule. Considering two real functions \(f(x)\) and \(g(x)\) that are continuous and differentiable on the interval \([a, b]\), such that \(f(a) = g(a) = 0\) and for all \(x \in (a, b), g'(x) \neq 0\), it holds that

\(\lim_{x \to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)}\)

provided the right-hand side limit exists.

Proof. Using Cauchy's theorem on the interval \((a, x)\) for \(x \in (a, b)\), we know that there exists \(c \in (a, x)\) such that

\(\frac{f'(c)}{g'(c)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f(x)}{g(x)}\).

Now, if we consider this expression in the limit of \(x \to a^+\), we can see that \(c \to a^+\) also, because \(c\) is always between \(a\) and \(x\). Hence:

\(\lim_{c \to a^+} \frac{f'(c)}{g'(c)} = \lim_{x \to a^+} \frac{f(x)}{g(x)}\),

which is what we wanted to prove. QED.

The theorem can similarly be proven for left-hand side limits (which then together form the l'Hôpital's rule we generally use). Hopefully I've made it clearer to you, if there's any questions feel free to ask.

Petar Veličković - 5 years, 3 months ago

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Simple and elegant proof!!!! Thanks!

Shubham Srivastava - 5 years, 3 months ago

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Here is how I think about it. Suppose you are watching two horses racing away from you, in the same direction, and the ratio of their speeds approaches, say, 2 as time goes on. Then you would expect that in the long run the faster horse will be about twice as far from you, compared to the slower one, regardless of what happens in the beginning of the race. Hopefully this makes sense to you.

Alexander Borisov - 5 years, 3 months ago

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Well, I was about to make a full proof, but Petar V. did it very well. So I'll just help in the intuitive part:

If you have two functions that runs to \(\infty\), how can you evaluate the limit of the ratio? Well, you need to know which function grows faster. The derivatives show us the growth of the function and the ratio of the derivatives, who grows faster.

\(x^2\) and \(x\) for exmeple, both goes to infinity. So what is the limit of \(\frac{x^2}{x}\)? Of course you can simplify, but you don't need to. You know that \(x^2\) grows much faster than \(x\), so it must be \(\infty\).

The idea behind \(\frac{0}{0}\) is equivalent.

Victor Chaves - 5 years, 3 months ago

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It makes intuitive sense. Let us think about the numerator and denominator of a number as pie and cutting it up. For example, 2/6 is two pies which are to be cut into 6 slices (not 6 slices each, just 6 slices in general). If we created a function for how many pies we have and how we are cutting it it makes intuitive sense that the rate at which we are making them divided by the rate at which we are cutting them is the same thing as our original process. It makes conceptual sense to me, it might not for you.

Alexander Sludds - 5 years, 3 months ago

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lim 1-cosmx/1-cosnx = ?
x to 0

Jayesh Chandan - 2 years, 7 months ago

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Well what stands for the logic part you may consider it as continued version of Stolz-Cesaro theorem.


Nicolae Sapoval - 5 years, 3 months ago

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