7} \frac{\sqrt{2+x}-3}{x-7}=|\frac{0}{0}|=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{2+x-9}=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{(\sqrt{2+x})^2-3^2}=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{(\sqrt{2+x}-3)(\sqrt{2+x}+3)}=\\\\lim_{x->7} \frac{1}{\sqrt{2+x}+3}=\frac{1}{\sqrt{2+7}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{3+3}=\\\\\frac{1}{6}" alt="lim_{x->7} \frac{\sqrt{2+x}-3}{x-7}=|\frac{0}{0}|=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{2+x-9}=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{(\sqrt{2+x})^2-3^2}=\\\\lim_{x->7} \frac{\sqrt{2+x}-3}{(\sqrt{2+x}-3)(\sqrt{2+x}+3)}=\\\\lim_{x->7} \frac{1}{\sqrt{2+x}+3}=\frac{1}{\sqrt{2+7}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{3+3}=\\\\\frac{1}{6}" align="absmiddle" class="latex-formula">
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\infty} \frac{-\frac{8}{x^3}+1}{\frac{18}{x^3}+1}=\frac{0+1}{0+1}=1" alt="lim_{x-\infty} \frac{-8+x^3}{18+x^3}=|\frac{\infty}{\infty}|=\\\\lim_{x->\infty} \frac{-\frac{8}{x^3}+1}{\frac{18}{x^3}+1}=\frac{0+1}{0+1}=1" align="absmiddle" class="latex-formula">