2} \frac{2x^2-2x^2}{4-x^2}=lim_{x->2} \frac{0}{4-x^2}=lim_{x->2} 0=0" alt="lim_{x->2} \frac{2x^2-2x^2}{4-x^2}=lim_{x->2} \frac{0}{4-x^2}=lim_{x->2} 0=0" align="absmiddle" class="latex-formula">
вложение
2} \frac{2x-x^2}{4-x^2}=lim_{x->2} \frac{x(2-x)}{(2-x)(2+x)}=lim_{x->2} \frac{x}{2+x}=\frac{2}{2+2}=0.5" alt="lim_{x->2} \frac{2x-x^2}{4-x^2}=lim_{x->2} \frac{x(2-x)}{(2-x)(2+x)}=lim_{x->2} \frac{x}{2+x}=\frac{2}{2+2}=0.5" align="absmiddle" class="latex-formula">
4} \frac{\sqrt{x}-2}{8-2x}=\\ lim_{x->4} \frac{\sqrt{x}-2}{2(4-x)}=\\ lim_{x->4} \frac{\sqrt{x}-2}{2(2-\sqrt{x})(2+\sqrt{x})}=\\ lim_{x->4} \frac{1}{-2(2+\sqrt{x})}=\\ \frac{1}{-2*(2+\sqrt{4})}=-0.125" alt="lim_{x->4} \frac{\sqrt{x}-2}{8-2x}=\\ lim_{x->4} \frac{\sqrt{x}-2}{2(4-x)}=\\ lim_{x->4} \frac{\sqrt{x}-2}{2(2-\sqrt{x})(2+\sqrt{x})}=\\ lim_{x->4} \frac{1}{-2(2+\sqrt{x})}=\\ \frac{1}{-2*(2+\sqrt{4})}=-0.125" align="absmiddle" class="latex-formula">
здесь не указано (предположил что x->0 - что дает повод обратиться к одной из замечательных границ)
0} \frac{5sin(\frac{1}{2}x)}{2x}=\\ lim_{x->0} \frac{5sin(\frac{1}{2}x)}{4*(\frac{1}{2}x)}=\frac{5*1}{4}=1.25" alt="lim_{x->0} \frac{5sin(\frac{1}{2}x)}{2x}=\\ lim_{x->0} \frac{5sin(\frac{1}{2}x)}{4*(\frac{1}{2}x)}=\frac{5*1}{4}=1.25" align="absmiddle" class="latex-formula">
0} \frac{\sqrt{1-x}-1}{x^2}=\\ lim_{x->0} \frac{(\sqrt{1-x}-1)(\sqrt{1-x}+1)}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{1-x-1}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{-x}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{-1}{x(\sqrt{1-x}+1)}=-\infty" alt="lim_{x->0} \frac{\sqrt{1-x}-1}{x^2}=\\ lim_{x->0} \frac{(\sqrt{1-x}-1)(\sqrt{1-x}+1)}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{1-x-1}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{-x}{x^2(\sqrt{1-x}+1)}=\\ lim_{x->0} \frac{-1}{x(\sqrt{1-x}+1)}=-\infty" align="absmiddle" class="latex-formula">