etudier la continuite, et derivabilite: x^2 sin((1/x)) si≠0 et f(0)=0


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Question Number 154337 by SANOGO last updated on 17/Sep/21

$e t u d i e r l a c o n t i n u i t e, e t d e r i v a b i l i t e :$ $x^{2} s i n (\frac{1}{x}) s i \neq 0 e t f (0) = 0$
	Answered by puissant last updated on 17/Sep/21
	$posons: { ((f(0)=0 si x=0 )),((f(x)=x^2 sin((1/x)) , sinon..)) :} ∗ Continuite^� : Premiere methode: lim_(x→0^+ ) f(x)= lim_(x→0^− ) f(x)=0 et f(0)=0 alors, f(x) est continu sur R.. Deuxieme methode: les fonctons carre^� (x^2 ), sin et inverse ((1/x)) sont continus sur R donc leur compose^� f(x) est continu sur R.. ∗ Derivabilite^� : on a : lim_(x→0) ((f(x)−f(0))/(x−0))=lim_(x→0) ((x^2 sin((1/x)))/x) =lim_(x→0) xsin((1/x)).. on a: −1≤sin((1/x))≤1 →−x≤xsin((1/x))≤x ⇒ lim_(x→0) xsin((1/x))=0 d′apres le theoreme des gendarmes.. D′ou f est derivable sur R^∗ ..$
	$p o s o n s : {\begin{cases} f (0) = 0 s i x = 0 \\ f (x) = x^{2} s i n (\frac{1}{x}), s i n o n . . \end{cases}$ $* C o n t i n u i t \overset{´}{e} :$ $P r e m i e r e m e t h o d e :$ $\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{-}} f (x) = 0 e t f (0) = 0$ $a l o r s, f (x) e s t c o n t i n u s u r R . .$ $D e u x i e m e m e t h o d e :$ $l e s f o n c t o n s c a r r \overset{´}{e} (x^{2}), s i n e t i n v e r s e (\frac{1}{x})$ $s o n t c o n t i n u s s u r R d o n c l e u r c o m p o s \overset{´}{e}$ $f (x) e s t c o n t i n u s u r R . .$ $* D e r i v a b i l i t \overset{´}{e} :$ $o n a : \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \lim_{x \to 0} \frac{x^{2} s i n (\frac{1}{x})}{x}$ $= \lim_{x \to 0} x s i n (\frac{1}{x}) . .$ $o n a : - 1 ⩽ s i n (\frac{1}{x}) ⩽ 1 \to - x ⩽ x s i n (\frac{1}{x}) ⩽ x$ $\Rightarrow \lim_{x \to 0} x s i n (\frac{1}{x}) = 0 d^{'} a p r e s l e t h e o r e m e$ $d e s g e n d a r m e s . .$ $D^{'} o u f e s t d e r i v a b l e s u r R^{*} . .$
	Commented by SANOGO last updated on 17/Sep/21

	$l e p u i s s a n t a p a r l e r m e r c i b i e n$
	Commented by puissant last updated on 17/Sep/21

	$D e r i e n . .$
	Answered by Jonathanwaweh last updated on 17/Sep/21
	$continuite lim_(x→0) x^2 sin((1/x))=0 car ∀x≠0 −1≤sin((1/x))≤1 donc −x^2 ≤x^2 sin((1/x))≤x^2 puisque { ((x^2 →o)),((−x^2 →0)) :}qd x→0 alors d′apres le theoreme de gendarme limx^2 sin((1/(x ))))=0 derivabilite on a lim_(x→0) ((f(x)−f(0))/(x−0))=lim_(x→0) xsin((1/x))=0 (applique tjr le theoreme des gendarmes) donc f est derivable en 0$
	$c o n t i n u i t e \lim_{x \to 0} x^{2} s i n (\frac{1}{x}) = 0 c a r \forall x \neq 0 - 1 ⩽ s i n (\frac{1}{x}) ⩽ 1$ $d o n c - x^{2} ⩽ x^{2} s i n (\frac{1}{x}) ⩽ x^{2} p u i s q u e {\begin{cases} x^{2} \to o \\ - x^{2} \to 0 \end{cases} q d x \to 0$ $a l o r s d^{'} a p r e s l e t h e o r e m e d e g e n d a r m e {l i m x}^{2} s i n (\frac{1}{x)}) = 0$ $d e r i v a b i l i t e$ $o n a \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \lim_{x \to 0} x s i n (\frac{1}{x}) = 0 (a p p l i q u e t j r l e t h e o r e m e d e s g e n d a r m e s)$ $d o n c f e s t d e r i v a b l e e n 0$
	Commented by SANOGO last updated on 17/Sep/21

	$m e r c i b i e n$
	Commented by puissant last updated on 17/Sep/21

	$B i e n M r j o n a t h a n . .$
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