Question

Номера на lim, 10 класс 26.30 и 26.31 с фотографии. Или хотябы 1, очень надо

Answer 1

26,30
а)
$\lim_{x \to 1} \frac{x^2+2x-3}{x-1} =\frac{(x-1)(x+3)}{x-1} =x+3=4$
б)
$\lim_{x \to -2} \frac{x+2}{2x^2+x-6} =\frac{x+2}{(x+2)(2x-3)} = \frac{1}{2x-3} = -\frac{1}{7}$
в)
$\lim_{x \to -1} \frac{x+1}{x^2-2x-3} =\frac{x+1}{(x-3)(x+1)} = \frac{1}{x-3} =- \frac{1}{4}$
г)
$\lim_{x \to 9} \frac{x^2-11x+18}{x-9} = \frac{(x-9)(x-2)}{x-9} =x-2=7$

26.31
а)
$\lim_{x \to -2} \frac{x+2}{x^3+8} =\frac{x+2}{(x+2)(x^2-2x+4)} = \frac{1}{x^2-2x+4} = \frac{1}{12}$
б)
$\lim_{x \to -1} \frac{1+x^3}{1-x^2} = \frac{(x+1)(x^2-x+1)}{(1-x)(1+x)} =\frac{(x^2-x+1)}{(1-x)} = \frac{3}{2}=1.5$
в)
$\lim_{x \to 3} \frac{x-3}{x^3-27} =\frac{x-3}{(x-3)(x^2+3x+9)} =\frac{1}{(x^2+3x+9)} = \frac{1}{27}$
г)
$\lim_{x \to 4} \frac{16-x^2}{64-x^3} = \frac{(4-x)(4+x)}{(4-x)(x^2+4x+16)} = \frac{(4+x)}{(x^2+4x+16)} = \frac{8}{48} = \frac{1}{6}$