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I have started learning limits in calculus and I came across this question:

Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ .

I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\dfrac{1}{x^5}}}{1+\dfrac{1}{x^3}} = \lim_{x \rightarrow \infty} \dfrac{\sqrt{9}}{1} = \boxed{3} $

But now I am asked to compute $ \displaystyle \lim_{x \rightarrow -\infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $

How to solve for minus infinity? Where to put minus sign , I am getting confused , please help. Thanks.

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Dividing by $|x^3|$ we get $$\frac{\sqrt{9x^6-x}}{x^3+1}=\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}$$ As $x\to-\infty$ we have, for $x<0$, \begin{align} \frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}&=\frac{\sqrt{\frac{9x^6-x}{x^6}}}{\frac{x^3+1}{-x^3}}=\frac{\sqrt{9-\frac{1}{x^5}}}{-1-\frac{1}{x^3}}\to\frac{\sqrt{9+0}}{-1+0}=\color{blue}{-3} \end{align}

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  • $\begingroup$ Oh , in this way we have to approach... Thank you very much , the solution has cleared my confusion :) $\endgroup$ – nsm Jan 15 '16 at 5:18
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Notice, one can easily change limit as $x\to +\infty$ as follows $$\lim_{x\to -\infty}\frac{\sqrt{9x^6-x}}{x^3+1}$$ $$=\lim_{x\to +\infty}\frac{\sqrt{9(-x)^6-(-x)}}{(-x)^3+1}$$

$$=\lim_{x\to +\infty}\frac{\sqrt{9x^6+x}}{1-x^3}$$ $$=\lim_{x\to +\infty}\frac{|3x^3|\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$ $$=\lim_{x\to +\infty}\frac{3x^3\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$ $$=3\lim_{x\to +\infty}\frac{\sqrt{1+\frac{1}{9x^5}}}{\frac{1}{x^3}-1}$$ $$=3\cdot \frac{\sqrt{1+0}}{0-1}=\color{red}{-3}$$

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You can also do $$\lim_{x \rightarrow -\infty} \frac{\sqrt{9x^6-x}}{x^3+1}$$ $$\sim \frac{3x^3}{-x^3} = -3$$
This is easier for me to do mentally to check my work; just realize that that the $9x^6$ term grows much faster than the $-x$ term towards postie infinity, so you effectively get $\sqrt{9x^6}$ as you go to infinity, and the addition on the bottom becomes insignificant while the cubic approaches negative infinity. Perhaps a little less rigid the way I word it, but conceptually simple.

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Notice:

$$\lim_{x\to-\infty}\frac{\sqrt{9x^6-x}}{x^3+1}=\lim_{x\to-\infty}\frac{\sqrt{9x^6-x}}{x^3}=$$ $$\lim_{x\to-\infty}-\sqrt{\frac{9x^6-x}{x^6}}=-\left(\lim_{x\to-\infty}\sqrt{\frac{9x^6-x}{x^6}}\right)=$$ $$-\left(\sqrt{\lim_{x\to-\infty}\frac{9x^6-x}{x^6}}\right)=-\left(\sqrt{\lim_{x\to-\infty}\frac{9-\frac{1}{x^5}}{1}}\right)=$$ $$-\sqrt{\frac{9-0}{1}}=-\sqrt{9}=-3$$

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  • $\begingroup$ This reproduces an approach already detailed in two earlier answers. What for? $\endgroup$ – Did Jan 17 '16 at 11:02

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