Граници 2

[donattelo]

Да се намерят границите:

[KOPMOPAH]

$\lim_{x \to 0}\frac{\sqrt{1+x^2}-1}{x^2}=\lim_{x \to 0}\frac{\sqrt{1+x^2}-1}{x^2}.\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1}=\\=\lim_{x \to 0}\frac{\left(\sqrt{1+x^2}\right)^2-1^2}{x^2\left(\sqrt{1+x^2}+1 \right)}=\lim_{x \to 0}\frac{1+x^2-1}{x^2\left(\sqrt{1+x^2}+1 \right)}=\\=\lim_{x \to 0}\frac{\cancel{1}+\cancel{x^2}-\cancel{1}}{\cancel{x^2}\left(\sqrt{1+x^2}+1 \right)}=\frac 0{\left(\sqrt{1+0}+1 \right)}=0$