\input /home/pasky/school/fastex/lib.tex

\subchapter{L'Hospital a p��tel�}{

\scpart{Idea}{
	Po��tat limity p�es derivace, kdy� jsou tak derivace definovan�.

	$$ \lim_{x\to0_+} {f(x) - f(a)\over g(x) - g(a)} = \lim {{f(x)-f(a)\over x-a}\over{g(x)-g(a)\over x-a}}
	   \hypothetically{=} {\lim\ldots\over\lim\ldots} \hypothetically{=} {f'(a)\over g'(a)} $$
}


\theorem{21}{Rolleova}{

	Nech� $f$ je spojit� na $[a,b]$, $a<b$ a nech� existuje $f'(x)$
	pro ka�d� $x \in (a,b)$. Nech� $f(a) = f(b)$. Pak existuje
	$\xi \in (a,b)$ takov�, �e $f'(\xi) = 0$.

	\proof{
	\list{
		\listItem{Nech� $f(x) = f(a)$ pro $\forall x \in [a,b]$, pak
			$f'(x) = 0$ pro $\forall x \in (a,b)$.
		}

		\listItem{Nech� $\existss x \in (a,b),\ f(x) \ne f(a)$,
			tedy bez �jmy na obecnosti $f(x)>f(a)$.
			Proto�e $f$ je spojit� na $[a,b]$, existuje
			lok�ln� extr�m $\xi \in [a,b]$ (nejv�t�� $f(x)$).
			Dle p�edpokladu $f'(\xi)$ mus� existovat, tedy
			podle \ias{}{20}{} je $f'(\xi) = 0$.
		}
	}

		\qed
	}
}


\theorem{22}{Lagrangeova v�ta o st�edn� hodnot�}{

	Nech� $f$ je spojit� na $[a,b]$, $a<b$
	a~nech� existuje $f'(x)$ pro $\forall x \in (a,b)$.
	Potom $\existss \xi \in (a,b)$ takov�, �e
	$$f'(\xi) = {f(b) - f(a) \over b - a}.$$
	(Tedy existuje v n�jak�m bod� derivace (te�na)
	rovnob�n� s naklon�nou podstavou
	(spojnic� krajn�ch bod�). \TODO{Obr�zek.})

	\proof{
		Definujme
		$$F(x) := f(x) - f(a) - {f(b) - f(a)\over b - a} \cdot (x-a)$$
		(na z�klad� divok� analyticko-geometrick� �vahy).

		Potom $F(a) = 0\ \land\ F(b) = 0$. Pak podle Rolleovy v�ty
		$\existss \xi \in (a,b): F'(\xi) = 0$. Tedy
		$$ 0 = F'(\xi) = f'(\xi) - {f(b)-f(a) \over b-a} $$
		a tud�� $$f'(\xi) = {f(b)-f(a)\over b-a}.$$

		\qed
	}

}


\theorem{23}{Cauchyova v�ta o st�edn� hodnot�}{

	Nech� $f,g$ jsou spojit� na $[a,b]$, $a<b$ a nech� $\existss f'(x)$ pro $\forall x \in (a,b)$
	a $\existss g'(x)$ vlastn� a nenulov� pro $\forall x \in (a,b)$. Potom $\existss \xi \in (a,b)$
	takov�, �e
	$$ {f(b)-f(a)\over g(b)-g(a)} = {f'(\xi) \over g'(\xi)} $$

	\proof{
		$g(a) \ne g(b)$, nebo� jinak by dle Rolleovy v�ty existovalo
		$\xi \in (a,b): g'(\xi) = 0$, co� by byl spor.

		Definujeme
		$$H(x) := \left(f(x)-f(a)\right)\left(g(x)-g(a)\right)
				- \left(g(x)-g(a)\right)\left(f(x)-f(a)\right).$$
		Potom $H(a) = H(b) = 0$, $H$ je spojit� na $[a,b]$, existuje $H'$ na $[a,b]$.

		Tedy dle Rolleovy v�ty $\existss \xi \in (a,b)$ takov�, �e
		$$0 = H'(\xi) = f'(\xi)\left(g\left(b\right)-g\left(a\right)\right)
				- g'(\xi)\left(f\left(b\right)-f\left(a\right)\right)
			\Rightarrow {f(b)-f(a)\over g(b)-g(a)} = {f'(\xi)\over g'(\xi)} $$
		(nebo� $g'(\xi)\ne0$, $g(b)-g(a) \ne 0$).

		\qed
	}

}


\theorem{24}{L'Hospitalovo pravidlo}{

	Nech� $a \in \real^*$ a funkce $f, g$ jsou definov�ny na n�jak�m $\cP^+(a,\delta)$.
	\list{
		\listItem{
			Nech� \M{\lim_{x\to a_+} f(x) = \lim_{a\to a_+} g(x) = 0}
			a existuje
			$$\lim_{x\to a_+} {f'(x)\over g'(x)}.$$
			Potom
			$$\lim_{x\to a_+} {f(x)\over g(x)} = \lim_{x\to a_+} {f'(x) \over g'(x)}.$$
		}

		\listItem{
			Nech� \M{\lim_{x\to a_+} |g(x)| = +\infty}
			a existuje
			$$\lim_{x\to a_+} {f'(x)\over g'(x)}.$$
			Potom
			$$\lim_{x\to a_+} {f(x)\over g(x)} = \lim_{x\to a_+} {f'(x) \over g'(x)}.$$
		}
	}

	\example{
		$$ \lim_{x\to 0} {x-\sin x \over x^3} = \lim_{x\to 0} {1-\cos x\over 3x^2}
			= \lim_{x\to 0} {\sin x \over 6x} = \lim_{x\to 0} {\cos x \over 6} = {1\over6} $$
	}

	\proof{
	\list{
	\namedListItem{i}{
		Nech� $a \in \real$ a nech�
		$$\lim_{x\to a_+} {f'(x)\over g'(x)} = A \in \real.$$
		Potom
		$$\existss \delta > 0,\ \forall x \in \cP^+(a,\delta):
			\ \existss f'(x) \in \real,\ g'(x) \in \real\setminus\{0\}.$$

		``Dodefinuji'' funkce $f$ a $g$ v bod� $A$: $f(a) = g(a) = 0$.
		Nyn� nech� $x \in \cP^+(a,\delta)$. (Obr�zek.) Potom $f$, $g$ jsou
		spojit� na $[a,x]$, nebo� ob� maj� vlastn� derivaci
		na cel�m $\cP^+(a,\delta)$.

		Tedy jsou spln�ny p�edpoklady \ias{}{23}{} (na intervalu $[a,x]$!)
		a proto plat�, �e $\existss \xi \in (a,x)$ takov�, �e
		$$ {f'(\xi) \over g'(\xi)} = {f(x)-f(a)\over g(x)-g(a)} $$
		($\xi$ z�vis� na $x$! Zna��me $\xi = \xi(x)$.)

		Zvolme $\eps > 0$. V�me, �e existuje $\omega > 0$ takov�, �e
		$$\forall y \in \cP^+(a,\omega): {f'(y) \over g'(y)} \in \cU(A,\eps).$$

		Pro $x \in \cP^+(a,\delta) \cap \cP^+(a,\omega)$ najdeme $\xi = \xi(x)$
		podle receptu na za��tku d�kazu. Potom
		$$ {f'(\xi) \over g'(\xi)} \in \cU(A,\eps). $$

		V�me ale, �e
		$$ {f(x)\over g(x)} = {f(x)-f(a) \over g(x)-g(a)} = {f'(\xi) \over g'(\xi)}, $$
		tedy tak�
		$$ {f(x)\over g(x)} \in \cU(A,\eps). $$

		Tak�e
		$$ \lim_{x\to a_+} {f(x)\over g(x)} = A = \lim_{x\to a_+} {f'(x)\over g'(x)}. $$
	}

	\namedListItem{i}{
		$a = -\infty$

		P�evedeme na p�edchoz� p��pad pomoc� substituce
		$$ \eqalign{F(y) = f\left(-{1\over y}\right),&\ F'(y) = f'\left(-{1\over y}\right) \cdot {1\over y}\cr
		            G(y) = g\left(-{1\over y}\right),&\ G'(y) = g'\left(-{1\over y}\right) \cdot {1\over y}}$$

		$$ \Longrightarrow \undernote{\lim_{y\to 0_+} {F(y)\over G(y)}}{= \lim_{x\to -\infty}{f(x)\over g(x)}}
					= \lim_{y\to 0_+} {F'(y) \over G'(y)}
					= \lim_{x\to -\infty} {f'(x) \over y'(x)} $$
	}

	\namedListItem{ii}{
		Technick� z�le�itost, pova�ujeme za dok�zan� (nezkou�� se).
	}
	}
	}

	\scpart{Pozn�mky, p��klady, varov�n�:}{
	\list{
		\listItem{\M{\lim_{x\to 0} {3x+1 \over 2x+2} \ne \lim_{x\to 0} {3\over2} = {2\over2}}}
		\listItem{\M{\lim_{x\to 0} {e^{-1/x} \over x} = \lim_{x\to 0} {e^{-1/x} \cdot {1\over x^2} \over 1}
			         = \lim_{x\to 0} {e^{-1/x} \over x^2}}}
		\listItem{\M{\lim_{x\to \infty} {x^2 \over x+\sin x} = +\infty}, ale
			\M{\lim_{x\to \infty} {2x \over 1 + \cos x}} neexistuje.}

		\listItem{L'Hospital vhodn� nap�. pro
			$$\lim_{x\to 0} {x-\sin x \over x^3},\ \lim_{x\to0} {e^x - x - 1 \over x^2},\ 
				\lim_{x\to 0} {\cos x - 1 + {x\over2}^2 \over x^4},\ \ldots$$
		}

		\listItem{\M{\lim_{x\to 1_-} (2-x)^{\tg\left(\pi x \over 2\right)}}
			$$ e^{\tg\left({\pi\over2}x\right){\log(2-x)\over1-x}(1-x)} = \tg({\pi\over2}x)(1-x)
				= {\sin({\pi\over2}x)\over\cos\left({\pi\over2}x\right)}(1-x) $$
			$$ \lim_{x\to 1_-} {1-x \over \cos\left({\pi\over2}x\right)}
				\becauseof{L'H}{=} \lim_{x\to 1_-}{-1 \over -\sin\left({\pi\over2}x\right)}{2\over\pi}
					\to {2\over\pi} $$
		}
		\listItem{\M{\lim_{n\to \infty} {n^2\over3^n} \becauseof{Heine + L'H ``${\infty\over\infty}$''}{=}
				\lim_{x\to \infty} {2x \over 3^x\log3} \becauseof{L'H ``${\infty\over\infty}$''}{=}
				\lim_{x\to \infty} {2 \over 3^x (\log 3)^2} = 0}}
		\listItem{\M{\lim_{x\to \infty} {\log x \over x} \becauseof{L'H ``${\infty\over\infty}$''}{=}
				\lim_{x\to \infty} {{1\over x}\over1} = 0}}
	}
	}

}

}

\bend
