Question

a) lim 3x+2/x^2-5x+4 ​

Answer 1

Ответ:

$1)\ \ \lim\limits _{x \to 0}\dfrac{3x+2}{x^2-5x+4}=\Big[\ \dfrac{0+2}{0-0+4}\ \Big]=\dfrac{2}{4}=\dfrac{1}{2}\\\\\\2)\ \ \lim\limits _{x \to 1}\dfrac{x^2-3x+2}{x^2-5x+4}=\lim\limits _{x \to 1}\dfrac{(x-1)(x-2)}{(x-1)(x-4)}=\lim\limits _{x \to 1}\dfrac{x-2}{x-4}=\dfrac{1-2}{1-4}=\dfrac{1}{3}\\\\\\\star \ x^2-3x+2=0\ \ ,\ \ x_1=1\ ,\ x_2=2\ \ (teorema\ Vieta)\ \star \\\\\star \ x^2-5x+4=0\ ,\ \ x_1=1\ ,\ x_2=4\ \ (teorema\ Vieta)\ \star$

$3)\ \ \lim\limits _{x \to \infty }\dfrac{4x^3-8x^2+5x-1}{x^3-3x+1}=\Big[\ \dfrac{:x^3}{:x^3}\ \Big]=\lim\limits _{x \to \infty}\dfrac{4-\frac{8}{x}+\frac{5}{x^2}-\frac{1}{x^3}}{1-\frac{3}{x^2}+\frac{1}{x^3}}=\\\\\\=\Big[\ \dfrac{4-0+0-0}{1-0+0}\ \Big]=4$