Bonjour
ClgPerdu
[tex]3)\ f(x)=\sqrt{4x^2+x}+2x\\\\\lim\limits_{x\to-\infty}f(x)=\lim\limits_{x\to-\infty}\dfrac{(\sqrt{4x^2+x}+2x)(\sqrt{4x^2+x}-2x)}{\sqrt{4x^2+x}-2x}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{(\sqrt{4x^2+x})^2-(2x)^2}{\sqrt{4x^2+x}-2x}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{4x^2+x-4x^2}{\sqrt{4x^2+x}-2x}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{x}{\sqrt{4x^2+x}-2x}[/tex]
[tex]=\lim\limits_{x\to-\infty}\dfrac{x}{\sqrt{4x^2(1+\dfrac{x}{4x^2})}-2x}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{x}{-2x\sqrt{1+\dfrac{1}{4x}}-2x}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{x}{-2x(\sqrt{1+\dfrac{1}{4x}}+1)}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{1}{-2(\sqrt{1+\dfrac{1}{4x}}+1)}\\\\\\=\dfrac{1}{-2(1+1)}\\\\\\=-\dfrac{1}{4}\\\\\Longrightarrow\boxed{\lim\limits_{x\to-\infty}f(x)=-\dfrac{1}{4}}[/tex]
[tex]4)\ f(x)=\dfrac{\sqrt{x^2-1}}{x-1}\\\\D_f=]-\infty;-1]\cup]1;+\infty[\\\\\\\boxed{\lim\limits_{x\to1^-}f(x)\ n'existe\ pas}\ car\ D_f=]-\infty;-1]\cup]1;+\infty[\\\\\\\lim\limits_{x\to1^+}f(x)=[\dfrac{0}{0}]=\lim\limits_{x\to1^+}\dfrac{\sqrt{(x-1)(x+1)}}{x-1}\\\\\\=\lim\limits_{x\to1^+}\dfrac{\sqrt{(x-1)(x+1)}\sqrt{x-1}}{(x-1)\sqrt{x-1}}\\\\\\=\lim\limits_{x\to1^+}\dfrac{\sqrt{(x-1)^2(x+1)}}{(x-1)\sqrt{x-1}}[/tex]
[tex]\\\\\\=\lim\limits_{x\to1^+}\dfrac{(x-1)\sqrt{x+1}}{(x-1)\sqrt{x-1}}\\\\\\=\lim\limits_{x\to1^+}\dfrac{\sqrt{x+1}}{\sqrt{x-1}}=[\dfrac{\sqrt{2}}{0^+}]\\\\\\=+\infty\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to1^+}f(x)=+\infty}[/tex]
[tex]\lim\limits_{x\to+\infty}f(x)=\lim\limits_{x\to+\infty}\dfrac{\sqrt{x^2(1-\dfrac{1}{x^2})}}{x(1-\dfrac{1}{x})}\\\\\\=\lim\limits_{x\to+\infty}\dfrac{x\sqrt{1-\dfrac{1}{x^2}}}{x(1-\dfrac{1}{x})}\\\\\\=\lim\limits_{x\to+\infty}\dfrac{\sqrt{1-\dfrac{1}{x^2}}}{1-\dfrac{1}{x}}=\dfrac{\sqrt{1-0}}{1-0}=\dfrac{1}{1}=1\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to+\infty}f(x)=1}[/tex]
[tex]\lim\limits_{x\to-\infty}f(x)=\lim\limits_{x\to-\infty}\dfrac{\sqrt{x^2(1-\dfrac{1}{x^2})}}{x(1-\dfrac{1}{x})}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{-x\sqrt{1-\dfrac{1}{x^2}}}{x(1-\dfrac{1}{x})}\\\\\\=\lim\limits_{x\to-\infty}\dfrac{-\sqrt{1-\dfrac{1}{x^2}}}{1-\dfrac{1}{x}}=\dfrac{-\sqrt{1-0}}{1-0}=\dfrac{-1}{1}=-1\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to-\infty}f(x)=-1}[/tex]
