
\begin{enumerate}
\item On pose $u(x)=\cos x$. Alors, $u'(x)=-\sin x$ et :
\[
\limite{x}{\pi}{\left(\dfrac{\cos x+1}{x-\pi}\right)}=\limite{x}{\pi}{\left(\dfrac{u(x)-u(\pi)}{x-\pi}\right)}=u'(\pi)=-\sin\pi=0.
\]

Ainsi, $\boxed{\limite{x}{\pi}{\dfrac{\cos x+1}{x-\pi}}=0}$

\item On pose $u(x)=\sqrt{x+1}$. Alors, $u'(x)=\dfrac{1}{2\sqrt{x+1}}$ et :
\[
\limite{x}{1}{\left( \dfrac{\sqrt{x+1}-\sqrt{2}}{x-1} \right) } = \limite{x}{1}{\left( \dfrac{u(x)-u(1)}{x-1} \right)}=u'(1)=\dfrac{1}{2\sqrt{2}}.
\]
Ainsi,
\[
\boxed{\limite{x}{1}{\dfrac{\sqrt{x+1}-\sqrt{2}}{x-1}}=\dfrac{1}{2\sqrt{2}}}
\]
\end{enumerate}
