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Question Number 188752 by Soma last updated on 06/Mar/23

Commented by Yacouba last updated on 06/Mar/23

$D f = {x \in R / x > 0 e t \ln x \neq 0}$
	Answered by a.lgnaoui last updated on 06/Mar/23
	$1•D_f ={x/x∈R^(∗+) et x≠1} D_f =]0,1[∪]1,+∞[ 2•limf(x)_(x→0^+ ) =lim_(x→0^+ ) (1−(1/(lnx))) =1−lim_(x→0^+ ) ((1/(lnx)) )=1 • limf(x)_(x→1^− ) =lim_(x→1^− ) ((−1)/(lnx))=+∞ •limf(x)_(x→1^+ ) =−∞ •limf(x)_(x→+∞) =lim_(x→+∞) ((x[(lnx+((lnx)/x))−(1/x)])/(lnx))= =lim((xlnx)/(lnx))=+∞ =lim((xlnx)/(lnx)) 3•lim(f(x)−x)_(x→+∞) =1−((1/(lnx)))=1 donc au voisinage de +∞:f(x)≅x+1 x+1 est donc Asymptote a la courbe de f(x). •x=0 x=1 (aussi asymptotes) 4•(f(x)−(x+1)=((−1)/(lnx))<0 donc l asymptote[est au dessous de courbe(C) 5•f^′ (x)=1−((1/(lnx)))^′ =1+(1/(x(lnx)^2 )) 6•x>0⇒f^′ (x)>0 :(f(x) croissante) 7•tengenteℸa (C) au point x=e est f(x)−f(e)=(x−e)f_(x=e) ^: (x) f(e)=e ⇒f′(e)=((e^2 +1)/e) (x+1−(1/(lnx)))−e=((e^2 +1)/e)(x−e) l equation est : x(1−((e^2 +1)/e))−(1/(lnx))+e^2 +1=0 8•Voir la courbe (C)$
	$1 ∙ D_{f} = {x / x \in R^{* +} e t x \neq 1}$ $D_{f} =] 0, 1 [\cup] 1, + \infty [$ $2 ∙ l i m f {(x)}_{x \to 0^{+}} = {l i m}_{x \to 0^{+}} (1 - \frac{1}{l n x})$ $= 1 - {l i m}_{x \to 0^{+}} (\frac{1}{l n x}) = 1$ $∙ l i m f {(x)}_{x \to 1^{-}} = {l i m}_{x \to 1^{-}} \frac{- 1}{l n x} = + \infty$ $∙ l i m f {(x)}_{x \to 1^{+}} = - \infty$ $∙ l i m f {(x)}_{x \to + \infty} = {l i m}_{x \to + \infty} \frac{x [(l n x + \frac{l n x}{x}) - \frac{1}{x}]}{l n x} =$ $= l i m \frac{x l n x}{l n x} = + \infty$ $= l i m \frac{x l n x}{l n x}$ $3 ∙ l i m {(f (x) - x)}_{x \to + \infty} = 1 - (\frac{1}{l n x}) = 1$ $d o n c a u v o i s i n a g e d e + \infty : f (x) ≅ x + 1$ $x + 1 e s t d o n c A s y m p t o t e a l a c o u r b e d e f (x) .$ $∙ x = 0 x = 1 (a u s s i a s y m p t o t e s)$ $4 ∙ (f (x) - (x + 1) = \frac{- 1}{l n x} < 0$ $d o n c l a s y m p t o t e [e s t a u d e s s o u s d e c o u r b e (C)$ $5 ∙ f^{^{'}} (x) = 1 - {(\frac{1}{l n x})}^{^{'}} = 1 + \frac{1}{x {(l n x)}^{2}}$ $6 ∙ x > 0 \Rightarrow f^{^{'}} (x) > 0 : (f (x) c r o i s s a n t e)$ $7 ∙ t e n g e n t e ℸ a (C) a u p o i n t x = e e s t$ $f (x) - f (e) = (x - e) f_{x = e}^{:} (x)$ $f (e) = e \Rightarrow f^{'} (e) = \frac{e^{2} + 1}{e}$ $(x + 1 - \frac{1}{l n x}) - e = \frac{e^{2} + 1}{e} (x - e)$ $l e q u a t i o n e s t :$ $x (1 - \frac{e^{2} + 1}{e}) - \frac{1}{l n x} + e^{2} + 1 = 0$ $8 ∙ V o i r l a c o u r b e (C)$
	Commented by a.lgnaoui last updated on 06/Mar/23

	Commented by a.lgnaoui last updated on 06/Mar/23

	$x : 0 1 + \infty$ $1 / + \infty ∣ - \infty / + \infty$
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