Question

Прошу решить примеры, в решении которых вы уверены.

Answer 1

$1) lim_{x to {-1}} frac{ x^{2} +(1-c)x-c}{ x^{2} +(2c+1)x+2c}= frac{0}{0}=lim_{x to {-1}} frac{ x^{2} +x-cx-c}{ x^{2} +2cx+x+2c}= \ = lim_{x to {-1}} frac{ x^{2} +x)-(cx+c)}{ (x^{2} +x)+(2cx+2c)}=lim_{x to {-1}} frac{ (x+1)(x-c)}{ (x+1)(x+2c)}= lim_{x to {-1}} frac{x-c}{x+2c}= \ = frac{-1-c}{-1+2c}$
При а=2; b=4; c=1
$2) lim_{x to infty} frac{ frac{4}{ x^{2}} - frac{2}{x} + frac{1}{x^{4} }+ frac{1}{x} }{frac{2}{ x^{2}} + frac{3}{x ^{4} } - frac{4}{x }- frac{1}{x} } = \ = lim_{x to infty} frac{ frac{1}{x^{4} }+frac{4}{ x^{2}} - frac{1}{x} }{ frac{3}{x ^{4} }+frac{2}{ x^{2}} - frac{5}{x }} = frac{0}{0}=$
Умножим и числитель и знаменатель на х:
$lim_{x to infty} frac{ frac{1}{x^{3} }+frac{4}{ x} -1 }{ frac{3}{x ^{3} }+frac{2}{ x}-5} = frac{-1}{-5}= frac{1}{5}=0,2$
$3) lim_{n to infty} frac{2n ^{4}-5n ^{2}+1}{4n+3n ^{2}-1 } =lim_{n to infty} frac{n ^{2} (2n ^{2}-5+ frac{1}{n ^{2} }) }{4n+3n ^{2}-1 } = \ =lim_{n to infty}n ^{2}cdot lim_{n to infty} frac{(2n ^{2}-5+ frac{1}{n ^{2} }) }{4n+3n ^{2}-1 } =infty cdot frac{2}{3} =infty$
$4) lim_{x to infty} (1+ frac{4}{x}+ frac{1}{x})^{2x}= lim_{x to infty} (1+ frac{5}{x})^{ frac{x}{5} 5cdot 2}= e ^{10}$

Answer 2

a =2 b=4 c=1