I have a problem with this limit question. $$\lim_{x \to \infty} \frac{x^3-4x}{7-2x^3}$$
How can the answer become $-\frac12$?
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3$\begingroup$ divide numerator and denominator by $x^3$ and note that as $x \to \infty$, $1/x \to 0$ $\endgroup$– AgentSCommented Sep 9, 2014 at 13:49
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$\begingroup$ If this doesn't make sense to you you should probably think about which terms "dominate" in the numerator and denominator as x gets big. The cubed terms dominate as x gets big, and then it's just the ratio of the coefficients. $\endgroup$– user420667Commented Sep 9, 2014 at 19:36
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4 Answers
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Factor the term with the highest power (here, $x^{3}$) :
$$ \begin{align*} \lim \limits_{x \to +\infty} \frac{x^{3}-4x}{7-2x^{3}} &= {} \lim \limits_{x \to + \infty} \frac{\require{cancel} \cancel{\color{blue}{x^{3}}} \big( 1 - \frac{4}{x^{2}} \big)}{\require{cancel} \cancel{\color{blue}{x^{3}}}\big( \frac{7}{x^{3}} - 2 \big)} \\[2mm] &= \lim \limits_{x \to +\infty} \frac{1 - \frac{4}{x^2}}{\frac{7}{x^3}-2} \\[2mm] &= \frac{1}{-2} \\[2mm] &= -\frac{1}{2} \end{align*} $$
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$$\frac{x^3-4x}{7-2x^3}=\frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2}$$
when $x\ne 0$. Now take the limit.
answered Sep 9, 2014 at 13:50
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Use L'hopital's rule, you have to derivate three times: $$\lim_{x\to \infty}(x^3-4x)/(7-2x^3)=\lim_{x\to \infty}(3x^2-4)/(-6x^2)=\lim_{x\to \infty}(6x)/(-12x)=\lim_{x\to \infty}(6)/(-12)=-1/2$$
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8$\begingroup$ This is the proverbial sledgehammer to crack a nut... $\endgroup$ Commented Sep 9, 2014 at 16:50
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1$\begingroup$ For this solution to be correct, you need to show that the original expression is an indeterminate form $\infty/\infty$, which is not completely obvious from the expression as written. $\endgroup$ Commented Sep 9, 2014 at 18:43
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Here are the steps $$ \lim_{x\to\infty} \frac{x^3-4x}{7-2x^3}= \lim_{x\to\infty} \frac{\frac{x^3}{x^3}-\frac{4x}{x^3}}{\frac{7}{x^3}-\frac{2x^3}{x^3}}= \lim_{x\to\infty} \frac{1-\frac{4}{x^2}}{\frac{7}{x^3}-2} =\frac{1-0}{0-2}=-\frac{1}{2} $$
