Question

40Б ПРЕДЕЛЫ функций решите

Answer 1

$1)\ \ \lim\limits_{x \to 2}\, \dfrac{4x^2-7x-2}{5x^2-9x-2}=\lim\limits_{x \to 2}\, \dfrac{4(x-2)(x+\frac{1}{4})}{5(x-2)(x+\frac{1}{5})}=\lim\limits_{x \to 2}\, \dfrac{4x+1}{5x+1}=\dfrac{8+1}{10+1}=\dfrac{9}{11}$

$2)\ \ \lim\limits_{x \to 2}\, \dfrac{x-2}{\sqrt{x+2}-2}=\lim\limits_{x \to 2}\, \dfrac{(x-2)(\sqrt{x+2}+2)}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}=\\\\\\=\lim\limits_{x \to 2}\, \dfrac{(x-2)(\sqrt{x+2}+2)}{(x+2)-4}=\lim\limits_{x \to 2}\, \dfrac{(x-2)(\sqrt{x+2}+2)}{x-2}=\\\\\\=\lim\limits_{x \to 2}\, (\sqrt{x+2}+2)=\sqrt{4}+2=2+2=4$

$3)\ \ \lim\limits\m_{x \to \infty}\, \dfrac{10x^2-3x+1}{x^2-8}=\lim\limits\m_{x \to \infty}\, \dfrac{10-\frac{3}{x}+\frac{1}{x^2}}{1-\frac{8}{x^2}}=\dfrac{10-0-0}{1-0}=\dfrac{10}{1}=10$

$4)\ \ \lim\limits_{x \to -5}\dfrac{x^3+125}{x+5}=\Big[\ \dfrac{0}{0}\ \Big]=\lim\limits_{x \to -5}\dfrac{(x+5)(x^2-5x+25)}{x+5}=\\\\\\=\lim\limits_{x \to -5}\, (x^2-5x+25)=(-5)^2-5\cdot (-5)+25=75$

$5)\ \ \lim\limits _{x \to 0}\, 40\cdot \dfrac{sin3x}{x}=\Big[\ \dfrac{0}{0}\ \Big]=40\cdot \lim\limits _{x \to 0}\Big(\underbrace{\dfrac{sin3x}{3x}}_{1}\cdot 3\Big)=40\cdot \lim\limits _{x \to 0}\ (1\cdot 3)=40\cdot 3=120\\\\\\\star\ \ \ \lim\limits _{\alpha (x) \to 0}\dfrac{sin\alpha (x)}{\alpha (x)}=1\ \ \ \ \ ,\ \ \alpha (x)=3x\ .\ \ \star$