Question

Распишите полностью пж

Answer 1

26.

а) $\lim_{x\to 2}\frac{x^{2}+2x-8}{8-x^{3}}=\lim_{x\to 2}\frac{x^{2}+4x-2x-8}{(2-x)(4+2x+x^{2})}=\lim_{x\to 2}\frac{x(x+4)-2(x+4)}{-(x-2)(4+2x+x^{2})}=\lim_{x\to 2}\frac{(x+4)-(x-2)}{-(x-2)(4+2x+x^{2})}=\lim_{x\to 2}\frac{(x+4)-(-1)}{4+2x+x^{2}}=\lim_{x\to 2}(-\frac{x+4}{x^{2}+2x+4})=-\frac{2+4}{2^{2}+2*2+4}=-\frac{6}{4+4+4}=-\frac{6}{12}=-\frac{1}{2}=-0,5$

б)

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