Answer
$+\infty$
Work Step by Step
$$\lim_{x \to \infty}\frac{x^{3}+1}{x^{2}\ln(x)}=\frac{ \infty}{ \infty}$$
Using the l'Hospital's rule it follows:
$$\lim_{x \to \infty}\frac{x^{3}+1}{x^{2}\ln(x)}=\lim_{x \to \infty}\frac{3x^{2}}{2x\ln(x)+x^{2}\frac{1}{x}}=\lim_{x \to \infty} \frac{3x^{2}}{2x\ln(x)+x}=\frac{ \infty}{ \infty}$$
Using the l'Hospital's rule second time it follows:
$$\lim_{x \to \infty} \frac{3x^{2}}{2x\ln(x)+x}=\lim_{x \to \infty} \frac{6x}{2\ln(x)+2+1}=\frac{ \infty}{ \infty}$$
Using the l'Hospital's rule thirdtime it follows:
$$\lim_{x \to \infty} \frac{6x}{2\ln(x)+2+1}=\lim_{x \to \infty} \frac{6}{2\frac{1}{x}}= \lim_{x \to \infty} 3x=+\infty$$
