Question

alguem consegue me ajudar?

Answer 1

Resposta:

Explicação passo a passo:

$\lim _{x\to \infty \:}\left(\frac{in\left(\frac{kx^2}{x^2+1}\right)^{9x}}{xf\left(x\right)}\right)\\explicacao\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)\\in\cdot \lim _{x\to \infty \:}\left(\frac{\left(\frac{kx^2}{x^2+1}\right)^{9x}}{xf\left(x\right)}\right)\\in\cdot \lim _{x\to \infty \:}\left(\frac{k^{9x}\left(x^2\right)^{9x}}{xf\left(x\right)\left(x^2+1\right)^{9x}}\right)$

$explicacao \lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0\\in\frac{\lim _{x\to \infty \:}\left(k^{9x}\left(x^2\right)^{9x}\right)}{\lim _{x\to \infty \:}\left(xf\left(x\right)\left(x^2+1\right)^{9x}\right)}$

    indinido

in---------------------  = 0

         $\infty$