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 $-\frac12$?
 A: Use L'hopital's rule, you have to derivate three times:
$$\lim_{x\to \infty}(x^3-4x)/(7-2x^3)=\lim_{x\to \infty}(3x^2-4)/(-6x^2)=\lim_{x\to \infty}(6x)/(-12x)=\lim_{x\to \infty}(6)/(-12)=-1/2$$
A: Factor the term with the highest power (here, $x^{3}$) :
$$ 
\begin{align*}
\lim \limits_{x \to +\infty} \frac{x^{3}-4x}{7-2x^{3}} &= {} \lim \limits_{x \to + \infty} \frac{\require{cancel} \cancel{\color{blue}{x^{3}}} \big( 1 - \frac{4}{x^{2}} \big)}{\require{cancel} \cancel{\color{blue}{x^{3}}}\big( \frac{7}{x^{3}} - 2 \big)} \\[2mm]
 &= \lim \limits_{x \to +\infty} \frac{1 - \frac{4}{x^2}}{\frac{7}{x^3}-2} \\[2mm]
 &= \frac{1}{-2} \\[2mm]
 &= -\frac{1}{2}
\end{align*}
$$
A: $$\frac{x^3-4x}{7-2x^3}=\frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2}$$
when $x\ne 0$. Now take the limit.
A: Here are the steps 
$$ \lim_{x\to\infty} \frac{x^3-4x}{7-2x^3}= \lim_{x\to\infty} \frac{\frac{x^3}{x^3}-\frac{4x}{x^3}}{\frac{7}{x^3}-\frac{2x^3}{x^3}}= \lim_{x\to\infty} \frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2} =\frac{1-0}{0-2}=-\frac{1}{2} $$
