\begin{enumerate}
\item Nous savons que $\limite{X}{+\infty}{\dfrac{\ln X}{X}}=0$.

\medskip

Posons $X=\sqrt{x^2-1}$. Alors, $\limite{x}{-\infty}{X}=+\infty$.

\medskip

De plus, $\dfrac{\ln\sqrt{x^2-1}}{x^2-1}=\dfrac{\ln X}{X^2}=\dfrac{\ln X}{X}\times\dfrac{1}{X}$.

\medskip

Ainsi, $\limite{x}{-\infty}{\left(\dfrac{\ln\sqrt{x^2-1}}{x^2-1}\right)}=\limite{X}{+\infty}{\left(\dfrac{\ln X}{X}\times\dfrac{1}{X}\right)}$. Or, $\limite{X}{+\infty}{\dfrac{1}{X}}=\limite{X}{+\infty}{\dfrac{\ln X}{X}}=0$. Ainsi,
\[
\boxed{\limite{x}{-\infty}{\left(\dfrac{\ln\sqrt{x^2-1}}{x^2-1} \right)}=0.}
\]

\item $\dfrac{\ln\left(x^2-2x+2 \right)}{(x-1)^2}=\dfrac{\ln\left[(x-1)^2+1 \right]}{(x-1)^2}=\dfrac{\ln(X+1)}{X}$, avec $X=(x-1)^2$.

\medskip

$\limite{x}{1}{X}=0$ donc $\limite{x}{1}{\left(\dfrac{\ln\left(x^2-2x+2 \right)}{(x-1)^2}\right)}=\limite{X}{0}{\dfrac{\ln(1+X)}{X}}=1$.\\
Ainsi,
\[
\boxed{\limite{x}{1}{\left(\dfrac{\ln\left(x^2-2x+2 \right)}{(x-1)^2}\right)}=1.}
\]

\item Posons $f(X)=\ln\left(1-X^2\right)$ et $g(x)=\ln\left(1+X\right)$, avec $X=\dfrac{1}{x}$.

Alors, $\limite{x}{+\infty}{X}=0$, et $f(0)=g(0)=\ln1=0$.
\begin{align*}
\dfrac{f(X)}{g(X)} & = \dfrac{f(X)-f(0)}{g(X)-g(0)}\\& = \dfrac{f(X)-f(0)}{X-0}\times\dfrac{X-0}{g(X)-g(0)}
\end{align*}
Par conséquent,
\begin{align*}
\limite{X}{0}{\dfrac{f(X)}{g(X)}} & = \limite{X}{0}{\left[\dfrac{f(X)-f(0)}{X-0}\times\dfrac{X-0}{g(X)-g(0)}\right]}
\end{align*}
Or, $\limite{X}{0}{\dfrac{f(X)-f(0)}{X-0}}=f'(0)$ et $\limite{X}{0}{\dfrac{X-0}{g(X)-g(0)}}=\dfrac{1}{g'(0)}$.

\medskip

$f'(X)=\dfrac{-2X}{1-X^2}$ et $g'(X)=\dfrac{1}{1+X}$.

Ainsi,
\[
\limite{X}{0}{\dfrac{f(X)}{g(X)}}=\dfrac{f'(0)}{g'(0)}=\dfrac{0}{1}=0
\]
et donc :
\[
\boxed{\limite{x}{+\infty}{\left(\dfrac{\ln\left(1-\dfrac{1}{x^2}\right)}{\ln\left(1+\dfrac{1}{x}\right)} \right)}=0.}
\]

\item On peut écrire :
\begin{align*}
\dfrac{\ln\left(1+\sqrt{x}\right)}{1-\sqrt{x+1}} & =\dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}\times\dfrac{\sqrt{x}}{1-\sqrt{x+1}}\times\dfrac{1+\sqrt{x+1}}{1+\sqrt{x+1}}\\
& = \dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}\times\dfrac{\sqrt{x}\left(1+\sqrt{x+1}\right)}{\textcolor{red}{1-(x+1)}}\\
& = \dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}\times\dfrac{\sqrt{x}\left(1+\sqrt{x+1}\right)}{\color{red}-x}\\
& = \dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}\times\dfrac{\cancel{\sqrt{x}}\left(1+\sqrt{x+1}\right)}{\color{red}-\sqrt x \times \cancel{\sqrt x}}\\
& = \dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}\times\left(-\dfrac{\left(1+\sqrt{x+1}\right)}{\sqrt{x}}\right)
\end{align*}

$\left.
\begin{array}{r}
\limite{x}{0}{\dfrac{\ln\left(1+\sqrt{x}\right)}{\sqrt{x}}}=\limite{X}{0}{\dfrac{\ln(1+X)}{X}}=1\\
\left.
\begin{array}{l}
\blacktriangleright\ \limite{x}{0}{\sqrt{x}}=0^+\\
\blacktriangleright\ \limite{X}{0^+}{\left(-\dfrac{1}{X}\right)}=-\infty\\
\blacktriangleright\ \limite{x}{0}{\left(1+\sqrt{x+1}\right)}=2
\end{array}\right\rbrace \Rightarrow \limite{x}{0}{\left(-\dfrac{\left(1+\sqrt{x+1}\right)}{\sqrt{x}}\right)}=-\infty
\end{array}\right\rbrace\Rightarrow\boxed{\limite{x}{0}{\left(\dfrac{\ln\left(1+\sqrt{x}\right)}{1-\sqrt{x+1}}\right)}=-\infty.}
$
\end{enumerate}