Inlocuind pe x cu 2 in limita obtinem cazul de nedeterminare [tex]\frac{0}{0}[/tex].
Deci putem aplica L'Hopital: derivam numaratorul si numitorul.
[tex]\lim_{x\to 2} \cfrac{(\sqrt{x^2+5}-\sqrt{x^2+4x-3})'}{(x^2-2x)'}=[/tex]
[tex]=\lim_{x\to 2}\cfrac{\cfrac{1}{2\sqrt{x^2+5}}(x^2+5)'-\cfrac{1}{2\sqrt{x^2+4x-3}}(x^2+4x-3)'}{2x-2}}=[/tex]
[tex]=\lim_{x\to 2}\cfrac{\cfrac{2x}{2\sqrt{x^2+5}}-\cfrac{2x+4}{2\sqrt{x^2+4x+3}}}{2x-2}=[/tex]
[tex]=\cfrac{\cfrac{4}{2\times3}-\cfrac{4+4}{2\times3}}{4-2}=\cfrac{4-8}{2\times6}=\cfrac{-4}{12}=-\cfrac{1}{3}[/tex]
