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[m]\lim_{x\to 3}\frac{3x^2-11x+6}{2x^2-5x-3}=(\frac{0}{0})=\lim_{x\to 3}\frac{(x-3)\cdot(3x-2)}{(x-3)\cdot(2x+1)}=\lim_{x\to 3}\frac{3x-2}
{2x+1}=[/m]
[m]=\frac{3\cdot 3-2}{2\cdot 3+1}=\frac{7}{7}=1[/m]
3)
[m]\lim_{x\to 5}\frac{x^2-8x+15}{x^2-25}=(\frac{0}{0})=\lim_{x\to 5}\frac{(x-5)\cdot(x-3)}{(x-5)\cdot(x+5)}=\lim_{x\to 3}\frac{x-3}{x+5}=\frac{5-3}{5+5}=[/m]
[m]=\frac{2}{10}[/m]
4)
[m]\lim_{x\to 3}\frac{3-x}{x^3-27}=(\frac{0}{0})=\lim_{x\to 3}\frac{-(x-3)}{(x-3)\cdot(x^2+x+9)}=\lim_{x\to 3}\frac{-1}{x^2+3x+9}=[/m]
[m]=\frac{-1}{3^2+3\cdot 3+9}=-\frac{1}{27}[/m]