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[latex]\\ \lim_{x\rightarrow\pi}{1-\cos{3x}\over ctgx}=\lim_{x\rightarrow\pi}{\sin x(1-\cos{3x})\over \cos x}={\sin \pi(1-\cos{3\pi})\over \cos \pi}={0*(1+1)\over-1}=0[/latex] [latex]\\[/latex] [latex]\\ \lim_{x\rightarrow \infty}{2x^2-5x+8\over3x^2+6x-15}=\lim_{x\rightarrow \infty}{2x^2\over3x^2}={2\over3}\\[/latex] [latex]\\[/latex] [latex]\\[/latex] [latex]\\ \lim_{x\rightarrow -2}{2x^2+7x+6 \over x^2+x-2}=\left | {0\over0} \right |=*\\\\2x^2+7x+6=0\\D=49-48=1\\x_1={-7+1\over4}={-3\over2}=-1,5\\x_2={-7-1\over4}=-2\\x^2+x-2=0\\x_1=-2, x_2=1\\\\*=\lim_{x\rightarrow -2}{(x+2)(2x+3)\over(x+2)(x-1)}=\lim_{x\rightarrow -2}{(2x+3)\over(x-1)}={-4+3\over-3}={1\over 3}[/latex] [latex]\\[/latex] [latex]\\\lim_{x\rightarrow 1}(3-2x^2)^{1\over2(1-x)}=*\\ \left [\lim_{x\rightarrow a}(1+u)^{1\over u}=e\\ \right ]\\ u=2-2x^2\\ {1\over u}={1\over{2-2x^2}}\\ \\*=\lim_{x\rightarrow 1}(1+(2-2x^2))^{{1\over2(1-x)}*{2(1-x^2)\over2(1-x^2)}}=\lim_{x\rightarrow 1}(1+(2-2x^2))^{{1+x\over2(1-x^2)}}=\lim_{x\rightarrow 1}e^{1+x}=e^2[/latex]