<s>[INST] How to solve this limits question I have a problem with this limit question.
$$\lim_{x \to \infty} \frac{x^3-4x}{7-2x^3}$$
How can the answer become $-\frac12$?
 [/INST] Here are the steps 
$$ \lim_{x\to\infty} \frac{x^3-4x}{7-2x^3}= \lim_{x\to\infty} \frac{\frac{x^3}{x^3}-\frac{4x}{x^3}}{\frac{7}{x^3}-\frac{2x^3}{x^3}}= \lim_{x\to\infty} \frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2} =\frac{1-0}{0-2}=-\frac{1}{2} $$
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