Question

lim -> 0 = x^4 + 2x^3+x^2 / x^2 + x

Answer 1

Explicação passo-a-passo:

Temos que:

$x^4+2x^3+x^2=(x^2+x)^2$

Assim:

$\lim_{x\to~0}~\dfrac{x^2+2x^3+x^2}{x^2+x}=\lim_{x\to~0}~\dfrac{(x^2+x)^2}{x^2+x}$

$\lim_{x\to~0}~\dfrac{x^2+2x^3+x^2}{x^2+x}=\lim_{x\to~0}~x^2+x$

$\lim_{x\to~0}~\dfrac{x^2+2x^3+x^2}{x^2+x}=0^2+0$

$\lim_{x\to~0}~\dfrac{x^2+2x^3+x^2}{x^2+x}=0$

Answer 2

Explicação passo-a-passo:

Olá!!

$lim \: \frac{{x}^{4} + 2 {x}^{3} + {x}^{2} }{ {x}^{2} + x} \\ x = > 0 \\ \\ lim \: \frac{{x}^{4} + {x}^{3} + {x}^{3} + {x}^{2} }{ x(x + 1)} \\ x = > 0 \\ \\ lim \: \frac{{x}^{3}(x + 1) + {x}^{2}(x + 1) }{ x(x + 1)} \\ x = > 0 \\ \\ lim \: \frac{({x}^{3} + {x}^{2}) \cancel{(x + 1) } }{ x \cancel{(x + 1)}} \\ x = > 0 \\ \\ lim \: \frac{({x}^{3} + {x}^{2}) }{ x } \\ x = > 0 \\ \\ lim \: \frac{ \cancel{x}({x}^{2} + x) }{ \cancel{x} } \\ x = > 0 \\ \\ lim \: {x}^{2} + x \\ x = > 0 \\ \\ substituindo \\ \\ {0}^{2} + 0 \\ \\ \checkmark\boxed{\boxed{Limite\Rightarrow0}}$