Question

ПОМОГИТЕ С ПРИМЕРОМ!!!!!!!

Answer 1

$1) \lim_{x \to 1} \frac{4x^2-3x-1}{5x-x^2-4}=?\\\\4x^2-3x-1=4(x-1)(x+0,25)\\D=(-3)^2-4*4*(-1)=9+16=25=5^2\\x_1=(3+5)/(2*4)=8/8=1\\x_2=(3-5)/(2*4)=-2/8=-0,25\\\\5x-x^2-4=-(x-1)(x-4)\\D=5^2-4*(-1)*(-4)=25-16=9=3^2\\x_1=(-5+3)/-2=-2/-2=1\\x_2=(-5-3)/-2=-8/-2=4 \\\\\lim_{x \to 1} \frac{4x^2-3x-1}{5x-x^2-4}= \lim_{x \to 1} \frac{4(x-1)(x+0,25)}{-(x-1)(x-4)}= \lim_{x \to 1} \frac{4x+1}{4-x}=\\\\= \frac{4*1+1}{4-1}= \frac{5}{3}=1 \frac{2}{3}$

$\lim_{x \to \infty} \frac{4x^2-3x-1}{5x-x^2-4}= \lim_{x \to \infty} \frac{ \frac{4x^2}{x^2} - \frac{3x}{x^2} - \frac{1}{x^2} }{ \frac{5x}{x^2} - \frac{x^2}{x^2} - \frac{4}{x^2} }= \\\\=\lim_{x \to \infty} \frac{ 4 - \frac{3}{x} - \frac{1}{x^2} }{ \frac{5}{x} - 1 - \frac{4}{x^2} }= \frac{4-0-0}{0-1-0}= \frac{4}{-1}=-4$