Resposta:
Explicação passo a passo:
[tex]\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)\\[/tex]
[tex]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)}\\[/tex]
indinido
in--------------------- = 0
[tex]\infty[/tex]
