L'Hôpital's Rule for Limits at Infinity

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Using L’Hôpital’s Rule, what is the value of $\lim_{x \to \infty}\frac{5x^3 - 4x^2 + 1}{7x^3 + 2x - 6}$ ?

$\frac{5}{3}$

$\frac{1}{7}$

$0$

$\frac{5}{7}$

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