5
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$$\lim_{x\to-\infty}\frac{\sqrt{x^6+729}}{4x^3+\sqrt{2x^6+1}}=\lim_{x\to-\infty}\frac{-\sqrt{1+\frac{729}{x^6}}}{4-\sqrt{2+\frac{1}{x^6}}}=\frac{-1}{4-\sqrt{2}}$$ $$\lim_{x\to\infty}\frac{\sqrt{x^6+729}}{4x^3+\sqrt{2x^6+1}}=\lim_{x\to-\infty}\frac{\sqrt{1+\frac{729}{x^6}}}{4+\sqrt{2+\frac{1}{x^6}}}=\frac{1}{4+\sqrt{2}}$$ Is this correct? Can someone please explain to me the distribution of the negatives in regards to the radicals when dealing with limits to $-\infty?$ Does it depend on the powers whether or not a negative stays outside of the radicals? Also, why does a negative not get applied to the $4$? Any help would be greatly appreciated!

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    Answers look good, remember if you do not like to work with a negative infinity in such limits, substitute $-x=t$ and convert the limit into a new one where t goes to pos infinity – 2017-02-24
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    Can confirm, answer looks fine to me – 2017-02-24
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    Can you please explain how the negatives distribute and why the negative does not apply to the $4$? @imranfat Maybe show me the answer how you do it with $-x=t$ please? – 2017-02-24

3 Answers 3

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Your result is correct.

About the sign of the radical for $ x \to -\infty$, remember that: $\sqrt{X^2}=|X|$, by definition, so, if $X<0$ we have $\sqrt{X^2}=-X$

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    hmm ok. And how come the $4$ does not get a negative? Because it is $4x^3$ why does the negative from $-\infty$ not stay with the 4? – 2017-02-24
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    Because $(4x^3)/x^3$ is still $4$ whatever the signs of $x$. – 2017-02-24
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    Thank You!! It just clicked.:) – 2017-02-24
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@Nick, here is how I did it:$$\lim_{t\to\infty}\frac{\sqrt{t^6+729}}{-4t^3+\sqrt{2t^6+1}}=\lim_{t\to\infty}\frac{\sqrt{1+\frac{729}{t^6}}}{-4+\sqrt{2+\frac{1}{t^6}}}=\frac{1}{-4+\sqrt{2}}$$

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    This might be studpid, but is $\frac{1}{-4+\sqrt{2}}=\frac{-1}{4-\sqrt{2}}$ – 2017-02-24
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    @NickPavini Yes it is the same. Signs have been switched. Also, type it in the calculator and compare... – 2017-02-24
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If $x\rightarrow -\infty$ then put $x=-\dfrac{1}{y}$ where $y>0$

If $x\rightarrow +\infty$ then put $x=\dfrac{1}{y}$ where $y>0$

Remember $\sqrt{x^2}=|x|$ and not $x$.