Questions mathématiques diverses

Question 34:
Déterminer la limite suivante :
$$\begin{align*} \lim\limits_{x \to 0}\dfrac{-1+\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+..\infty}}}}{-1+\sqrt{x^3+\sqrt{x^3+\sqrt{x^3+\sqrt{x^3+...\infty}}}}} \end{align*}$$

Réponse 34:
Posons $$\begin{align*} &y=\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+..\infty}}}\\ &z=\sqrt{x^3+\sqrt{x^3+\sqrt{x^3+...\infty}}} \end{align*}$$ Alors $$\begin{align*} &🔶y=\sqrt{(\tan x-\sin x)+y}\\ &\iff y^2=(\tan x-\sin x)+y\;et\;y\geqslant0\\ &\iff y^2-y+(\tan x-\sin x)=0\;et\;y\geqslant0\\ &y=\dfrac{1+\sqrt{1+4(\tan x-\sin x)}}{2}\\ &🔶z=\sqrt{x^3+z}\\ &\iff z^2=x^3+z\;et\;z\geqslant0\\ &\iff z^2-z-x^3=0\;et\;z\geqslant0\\ &\iff z=\dfrac{1+\sqrt{1+4x^3}}{2}\\ &🔶\dfrac{-1+y}{-1+z}=\dfrac{-1+\dfrac{1+\sqrt{1+4(\tan x-\sin x)}}{2}}{-1+\dfrac{1+\sqrt{1+4x^3}}{2}}\\ &=\dfrac{\sqrt{1+4(\tan x-\sin x)}-1}{\sqrt{1+4x^3}-1}\\ &=\dfrac{4(\tan x-\sin x)}{4x^3}\dfrac{\sqrt{1+4x^3}+1}{\sqrt{1+4(\tan x-\sin x)}+1}\\ &Donc\\ &\lim\limits_{x \to 0}\dfrac{-1+y}{-1+z}=\lim\limits_{x \to 0}\dfrac{4(\tan x-\sin x)}{4x^3}\times\dfrac{2}{2}\\ &\qquad\qquad=\lim\limits_{x \to 0}\dfrac{\sin x(1-\cos x)}{x^3\cos x}\\ &\qquad\qquad=\lim\limits_{x \to 0}\dfrac{\sin x}{x}\dfrac{1-\cos x}{x^2}\dfrac{1}{\cos x}\\ &\qquad\qquad=1\times\dfrac{1}{2}\times1=\dfrac{1}{2} \end{align*}$$