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<div class="ccc"> 
<ol><li> On a, en 
!insmath 1
, 
<center>
!insmath  x-1 = O(\ln x), \quad x^3-5x+4 = O(x^2 + 6x -7).
  
</center>  
</li><li>
 On a, à l'infini,
<center> 
!insmath  x^3 - 5 x + 4 = O(99 x^3 -5), \quad \ln (x^{1000}+1) = O(\ln x).
  
</center>  </li></ol>
</div>
