
$\begin{aligned}[t]
f(x) & = \dfrac{\sqrt{x^2+x+2}-2}{x-1}\times\dfrac{\sqrt{x^2+x+2}+2}{\sqrt{x^2+x+2}+2}\\
& = \dfrac{x^2+x-2}{(x-1)\left(\sqrt{x^2+x+2}+2\right)}
\end{aligned}$

$\begin{aligned}[t]
f(x) & = \dfrac{(x-1)(x+2)}{(x-1)\left(\sqrt{x^2+x+2}+2\right)}\\
&= \dfrac{x+2}{\sqrt{x^2+x+2}+2} \quad\text{si }x\neq1
\end{aligned}$

\medskip

Ainsi, $\limite{x}{1}{f(x)}=\dfrac{3}{4}=f(1)$. Donc $f$ est continue en 1.
