Bonsoir Carolinaaaaa
[tex]1)\ \lim\limits_{x\to0}\dfrac{\sqrt{1+x}-1}{x}=[\dfrac{0}{0}]=\lim\limits_{x\to0}\dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}\\\\\\=\lim\limits_{x\to0}\dfrac{(\sqrt{1+x})^2-1^2}{x(\sqrt{1+x}+1)}=\lim\limits_{x\to0}\dfrac{1+x-1}{x(\sqrt{1+x}+1)}=\lim\limits_{x\to0}\dfrac{x}{x(\sqrt{1+x}+1)}\\\\\\=\lim\limits_{x\to0}\dfrac{1}{\sqrt{1+x}+1}=\dfrac{1}{\sqrt{1+0}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to0}\dfrac{\sqrt{1+x}-1}{x}=\dfrac{1}{2}}[/tex]
[tex]2)\ \lim\limits_{x\to+\infty}\left(\sqrt{x^2+1}-x\right)=[+\infty-\infty]\\\\\\=\lim\limits_{x\to+\infty}\dfrac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}=\lim\limits_{x\to+\infty}\dfrac{\left(\sqrt{x^2+1}\right)^2-x^2}{\sqrt{x^2+1}+x}\\\\\\=\lim\limits_{x\to+\infty}\dfrac{(x^2+1)^2-x^2}{\sqrt{x^2+1}+x}=\lim\limits_{x\to+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\\\\\\=\lim\limits_{x\to+\infty}\dfrac{1}{\sqrt{x^2+1}+x}=[\dfrac{1}{+\infty}]=0[/tex]
[tex]\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to+\infty}\left(\sqrt{x^2+1}-x\right)=0}[/tex]
[tex]3)\ \lim\limits_{x\to0}\dfrac{x^2}{1-\sqrt{1-x^2}}=[\dfrac{0}{0}]=\lim\limits_{x\to0}\dfrac{x^2(1+\sqrt{1-x^2})}{(1-\sqrt{1-x^2})(1+\sqrt{1-x^2})}\\\\\\=\lim\limits_{x\to0}\dfrac{x^2(1+\sqrt{1-x^2})}{1^2-(\sqrt{1-x^2})^2}=\lim\limits_{x\to0}\dfrac{x^2(1+\sqrt{1-x^2})}{1-(1-x^2)}=\lim\limits_{x\to0}\dfrac{x^2(1+\sqrt{1-x^2})}{1-1+x^2}\\\\\\=\lim\limits_{x\to0}\dfrac{x^2(1+\sqrt{1-x^2})}{x^2}=\lim\limits_{x\to0}\dfrac{1+\sqrt{1-x^2}}{1}=\lim\limits_{x\to0}\left(1+\sqrt{1-x^2}\right)\\\\\\=1+\sqrt{1-0}=1+1=2[/tex]
[tex]\Longrightarrow\boxed{\lim\limits_{x\to0}\dfrac{x^2}{1-\sqrt{1-x^2}}=2}[/tex]
[tex]4)\ \lim\limits_{x\to1^-}\dfrac{\left \lfloor x \right \rfloor}{x}=\dfrac{0}{1}=0\\\\\\\lim\limits_{x\to1^+}\dfrac{\left \lfloor x \right \rfloor}{x}=\dfrac{1}{1}=1\\\\\\\Longrightarrow\boxed{\lim\limits_{x\to1}\dfrac{\left \lfloor x \right \rfloor}{x}\ \text{n'existe pas}}[/tex]
