[tex]cos(x)=1-2sin^2(\frac{x}{2})[/tex]
Inlocuim in limita:
[tex]\lim_{x \to 0} \frac{1-(1-2sin^2(\frac{x}{2}))}{x(\sqrt{1+x}-1)}=\lim_{x\to 0} \frac{2sin^2(\frac{x}{2})}{x(\sqrt{1+x}-1)}[/tex]
Dar cunoastem limita:
[tex]\lim_{x\to 0} \frac{sin(x)}{x}=1[/tex]
Astfel ca putem inmulti si imparti limita noastra cu [tex](\frac{x}{2})^2[/tex] pentru a obtine:
[tex]\lim_{x\to 0} \frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2x(\sqrt{1+x}-1)}\times(\frac{x}{2})^2=\lim_{x\to0} (\frac{sin(\frac{x}{2})}{\frac{x}{2}})^2\times \frac{2x^2}{4x(\sqrt{1+x}-1)}=[/tex]
[tex]= 1^2\times \lim_{x\to0} \frac{x}{2(\sqrt{1+x}-1)}=\lim_{x\to0} \frac{x}{2(\sqrt{1+x}-1)}[/tex]
Acum putem inmulti si imparti cu conjugatul parantezei de la numitor si vom obtine:
[tex]\lim_{x\to0} \frac{x(\sqrt{1+x}+1)}{2(\sqrt{1+x}-1)(\sqrt{1+x}+1)}=\lim_{x\to0} \frac{x(\sqrt{1+x}+1)}{2(\sqrt{1+x}^2-1^2)}=\lim_{x\to0} \frac{x(\sqrt{1+x}+1)}{2(1+x-1)}=[/tex]
[tex]=\lim_{x\to0}\frac{x(\sqrt{1+x}+1)}{2x}=\lim_{x\to0} \frac{\sqrt{1+x}+1}{2}=\frac{2}{2}=1[/tex]
