Высшая математика! Найти приделы ф-ции. С полным ходом решения пожалуйста.
Найти пределы функций:0}\frac{\sqrt{2-x}-\sqrt{2}}{x}; \lim_{x->3}\frac{9-x^{2}}{\sqrt{3x}-3}; \lim_{x->5}\frac{2-\sqrt{x-1}}{x^{2}-25}; \lim_{x->6}\frac{\sqrt{x-2}-2}{x-6};" alt="\lim_{x->0}\frac{\sqrt{2-x}-\sqrt{2}}{x}; \lim_{x->3}\frac{9-x^{2}}{\sqrt{3x}-3}; \lim_{x->5}\frac{2-\sqrt{x-1}}{x^{2}-25}; \lim_{x->6}\frac{\sqrt{x-2}-2}{x-6};" align="absmiddle" class="latex-formula"> 0}\frac{\sqrt{1+4x}-1}{x}" alt="\lim_{x->0}\frac{\sqrt{1+4x}-1}{x}" align="absmiddle" class="latex-formula"> Решение: 0}\frac{\sqrt{2-x}-\sqrt{2}}{x}=\lim_{x->0}\frac{(\sqrt{2-x}-\sqrt{2})(\sqrt{2-x}+\sqrt{2})}{x(\sqrt{2-x}+\sqrt{2})}=" alt="\lim_{x->0}\frac{\sqrt{2-x}-\sqrt{2}}{x}=\lim_{x->0}\frac{(\sqrt{2-x}-\sqrt{2})(\sqrt{2-x}+\sqrt{2})}{x(\sqrt{2-x}+\sqrt{2})}=" align="absmiddle" class="latex-formula"> 0}\frac{2-x-2}{x(\sqrt{2-x}+\sqrt{2})}=\lim_{x->0}\frac{-x}{x(\sqrt{2-x}+\sqrt{2})}=-\lim_{x->0}\frac{1}{\sqrt{2-x}+\sqrt{2}}=" alt="=\lim_{x->0}\frac{2-x-2}{x(\sqrt{2-x}+\sqrt{2})}=\lim_{x->0}\frac{-x}{x(\sqrt{2-x}+\sqrt{2})}=-\lim_{x->0}\frac{1}{\sqrt{2-x}+\sqrt{2}}=" align="absmiddle" class="latex-formula"> 3}\frac{9-x^{2}}{\sqrt{3x}-3}=\lim_{x->3}\frac{(9-x^{2})(\sqrt{3x}+3)}{(\sqrt{3x}-3)(\sqrt{3x}+3)}=" alt="\lim_{x->3}\frac{9-x^{2}}{\sqrt{3x}-3}=\lim_{x->3}\frac{(9-x^{2})(\sqrt{3x}+3)}{(\sqrt{3x}-3)(\sqrt{3x}+3)}=" align="absmiddle" class="latex-formula"> 3}\frac{(9-x^{2})(\sqrt{3x}+3)}{3x-9}=\lim_{x->3}\frac{(3-x)(3+x))(\sqrt{3x}+3)}{3(x-3)}=" alt="=\lim_{x->3}\frac{(9-x^{2})(\sqrt{3x}+3)}{3x-9}=\lim_{x->3}\frac{(3-x)(3+x))(\sqrt{3x}+3)}{3(x-3)}=" align="absmiddle" class="latex-formula"> 3}\frac{(3+x)(\sqrt{3x}+3)}{3}=-\frac{(3+3)(\sqrt{3*3}+3))}{3}=-12" alt="=-\lim_{x->3}\frac{(3+x)(\sqrt{3x}+3)}{3}=-\frac{(3+3)(\sqrt{3*3}+3))}{3}=-12" align="absmiddle" class="latex-formula"> 5}\frac{2-\sqrt{x-1}}{x^{2}-25}=\lim_{x->5}\frac{(2-\sqrt{x-1})(2+\sqrt{x-1})}{(x^{2}-25)(2+\sqrt{x-1})}=" alt="\lim_{x->5}\frac{2-\sqrt{x-1}}{x^{2}-25}=\lim_{x->5}\frac{(2-\sqrt{x-1})(2+\sqrt{x-1})}{(x^{2}-25)(2+\sqrt{x-1})}=" align="absmiddle" class="latex-formula"> 5}\frac{4-x+1}{(x^{2}-25)(2+\sqrt{x-1})}=\lim_{x->5}\frac{5-x}{(x-5)(x+5)(2+\sqrt{x-1})}=" alt="=\lim_{x->5}\frac{4-x+1}{(x^{2}-25)(2+\sqrt{x-1})}=\lim_{x->5}\frac{5-x}{(x-5)(x+5)(2+\sqrt{x-1})}=" align="absmiddle" class="latex-formula"> 5}\frac{1}{(x+5)(2+\sqrt{x-1})}=-\frac{1}{(5+5)(2+\sqrt{5-1})}=-\frac{1}{40}" alt="=-\lim_{x->5}\frac{1}{(x+5)(2+\sqrt{x-1})}=-\frac{1}{(5+5)(2+\sqrt{5-1})}=-\frac{1}{40}" align="absmiddle" class="latex-formula"> 6}\frac{\sqrt{x-2}-2}{x-6}=\lim_{x->6}\frac{(\sqrt{x-2}-2)(\sqrt{x-2}+2)}{(x-6)(\sqrt{x-2}+2)}=\lim_{x->6}\frac{x-2-4}{(x-6)(\sqrt{x-2}+2)}" alt="\lim_{x->6}\frac{\sqrt{x-2}-2}{x-6}=\lim_{x->6}\frac{(\sqrt{x-2}-2)(\sqrt{x-2}+2)}{(x-6)(\sqrt{x-2}+2)}=\lim_{x->6}\frac{x-2-4}{(x-6)(\sqrt{x-2}+2)}" align="absmiddle" class="latex-formula"> 6}\frac{x-6}{(x-6)(\sqrt{x-2}+2)}=\lim_{x->6}\frac{1}{\sqrt{x-2}+2}=\frac{1}{\sqrt{6-2}+2}=\frac{1}{2+2}=\frac{1}{4}" alt="=\lim_{x->6}\frac{x-6}{(x-6)(\sqrt{x-2}+2)}=\lim_{x->6}\frac{1}{\sqrt{x-2}+2}=\frac{1}{\sqrt{6-2}+2}=\frac{1}{2+2}=\frac{1}{4}" align="absmiddle" class="latex-formula"> 0}\frac{\sqrt{1+4x}-1}{x}=\lim_{x->0}\frac{(\sqrt{1+4x}-1)(\sqrt{1+4x}+1)}{x(\sqrt{1+4x}+1)}=\lim_{x->0}\frac{1+4x-1}{x(\sqrt{1+4x}+1)}" alt="\lim_{x->0}\frac{\sqrt{1+4x}-1}{x}=\lim_{x->0}\frac{(\sqrt{1+4x}-1)(\sqrt{1+4x}+1)}{x(\sqrt{1+4x}+1)}=\lim_{x->0}\frac{1+4x-1}{x(\sqrt{1+4x}+1)}" align="absmiddle" class="latex-formula"> 0}\frac{4x}{x(\sqrt{1+4x}+1)}=\lim_{x->0}\frac{4}{\sqrt{1+4x}+1}=\frac{4}{\sqrt{1+4*0}+1}=2" alt="=\lim_{x->0}\frac{4x}{x(\sqrt{1+4x}+1)}=\lim_{x->0}\frac{4}{\sqrt{1+4x}+1}=\frac{4}{\sqrt{1+4*0}+1}=2" align="absmiddle" class="latex-formula">
Очень помогли, спасибо огромное. Выручили)