Question

Решите придел(lim), plz.

Answer 1

$displaystyle lim_{x to infty} (frac{x-2}{x+1})^{2x-3}=(frac{infty}{infty})^infty=lim_{x to infty}(frac{x+1-3}{x+1})^{2x-3}=\=[lim_{x to infty}(1+frac{1}{-frac{x+1}{3}})^{-frac{x+1}{3}}}]^{-frac{3}{x+1}*(2x-3)}=e^{displaystyle lim_{x to infty}frac{9-6x}{x+1}}=\=e^{displaystyle lim_{x to infty}frac{x(frac{9}{x}-6)}{x(1+frac{1}{x})}}=e^{-6}=frac{1}{e^6}$

Answer 2

(x-2/x+1)

Answer 3

а тут сколько?

Answer 4

x1