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$$\lim_{x\to-\infty}\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}$$ This is what I have tried so far, $$\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10}+x^2)^{\frac{1}{2}}}$$

$$\begin{align} &\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{10})^{\frac{1}{2}}+(x)^{\frac{1}{2}}}\\ &\lim_{x\to-\infty}\frac{2x^5+x^2}{(2x^{5})+(x)}\\ &\lim_{x\to-\infty}\frac{x^2}{x}\\ &\lim_{x\to-\infty}x = -\infty\\ \end{align}$$ Is this the correct approach to finding the limit?

Asaf Karagila
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Kot
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2 Answers2

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Observe that for $x<0$, we have $$\begin{align}\sqrt{2x^{10}+x^2} &= \left(\sqrt{x^2}\right)^5\sqrt{2+\frac1{x^8}}\\ &= |x|^5\sqrt{2+\frac1{x^8}}\\ &= -x^5\sqrt{2+\frac1{x^8}}.\end{align}$$ Does that get you started?

Let me approach this two other ways, rigorously, and loosely. Note that the function is undefined for $x=0$, so we will only consider $x\neq 0$ in the following. To study the end behavior of this function, we will ultimately be interested only in the terms of highest degree on top and on bottom. Ideally, we'd like to rewrite the quotient in an equivalent way so that at least one of the highest degree terms is constant. Once we've done that, we'll proceed to the limit. I will call the function $f(x).$

Rigorous:

You expressed a desire to divide top and bottom by $x^5,$ so I will do it that way. The key fact that I will be using several times is that for any real $\alpha,$ we have $$|\alpha|=\sqrt{\alpha^2}.$$ The following manipulations hold for any $x\neq 0:$ $$\begin{align}f(x) &= \frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}}{x^{-5}}\cdot\frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}}\\ &= \frac{x^{-5}(2x^5+x^2)}{x^{-5}\sqrt{2x^{10}+x^2}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2(2x^8+1)}}\\ &= \frac{2+x^{-3}}{x^{-5}\sqrt{x^2}\sqrt{2x^8+1}}\\ &= \frac{2+x^{-3}}{x^{-5}|x|\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}.\end{align}$$ Observe that $x^{-4}$ is positive for all $x\neq 0,$ so in particular, $$x^{-4}=\left|x^{-4}\right|=\sqrt{(x^{-4})^2}=\sqrt{x^{-8}}.$$

Hence, we have $$\begin{align}f(x) &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{x^{-4}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}}\sqrt{2x^8+1}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{x^{-8}\left(2x^8+1\right)}}\\ &= \frac{x}{|x|}\cdot\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\end{align}$$ for all $x\neq 0$. Now, note that $$\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}=\sqrt{2},$$ as I believe you've already calculated. Also note that $$\frac{x}{|x|}=\begin{cases}1 & \text{if }x>0,\\-1 & \text{if }x<0,\end{cases}$$ and so $$\lim_{x\to-\infty}\frac{x}{|x|}=-1.$$ Therefore, $$\lim_{x\to-\infty}f(x)=\left[\lim_{x\to-\infty}\frac{x}{|x|}\right]\cdot\left[\lim_{x\to-\infty}\frac{2+x^{-3}}{\sqrt{2+x^{-8}}}\right]=-1\cdot\sqrt{2}=-\sqrt{2}.$$

Loose:

This gets back to what Babak S. mentions in the comment below. When dealing with end-behavior of polynomials, only the highest-degree term ultimately matters. To that end, we can (roughly speaking) "drop" all the terms in numerator and denominator except those of highest degree, and then find the limit that way. That is, $$\begin{align}\lim_{x\to-\infty}f(x) &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2x^{10}}}\\ &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2}\sqrt{x^{10}}}\\ &= \frac{2}{\sqrt{2}}\cdot\lim_{x\to-\infty}\frac{x^5}{\sqrt{(x^5)^2}}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}\frac{x^5}{|x^5|}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}-1\\ &= -\sqrt{2}.\end{align}$$

Cameron Buie
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  • I am trying to follow along. In the end, the absolute value becomes negative because x is approaching negative infinity or that $x<0$? – Kot Feb 27 '13 at 02:39
  • Both, actually. As $x$ approaches $-\infty$, we will eventually have $x<0$, at which point we will have $|x|=-x$. – Cameron Buie Feb 27 '13 at 02:42
  • @StevenN: Note that if you have $a_nx^n+...+a_1x+a_0$ and $x$ goes to $\infty$ then $$a_nx^n+...+a_1x+a_0\sim a_nx^n$$ So $\sqrt{2x^{10}+x^2}\sim\sqrt{2}|x^5|=\sqrt{2}\times(-x^5)$ since $x\to-\infty$ – Mikasa Feb 27 '13 at 02:43
  • Nice, Cameron! +1 – amWhy Feb 27 '13 at 02:45
  • So would the limit be $\frac{2}{\sqrt{2}}$? – Kot Feb 27 '13 at 02:57
  • @StevenN: Close. You accidentally changed signs somewhere. It should be $$\frac{2}{-\sqrt 2}=-\frac{2}{\sqrt 2}=-\sqrt{2}.$$ – Cameron Buie Feb 27 '13 at 03:30
  • @CameronBuie I changed the signs because it was $-x^5$, as x approaches negative infinity, wouldn't it become positive since the two negatives cancel out? – Kot Feb 27 '13 at 03:32
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    @StevenN: Ah! I see what happened. The next step would be to rewrite $$\cfrac{2x^5+x^2}{-x^5\sqrt{2+\frac1{x^8}}}=\cfrac{2+\frac1{x^3}}{-\sqrt{2+\frac{1}{x^8}}},$$ so while it's true that $-x^5$ is positive, $x^5$ is negative, and so their quotient leaves the $-$ sign in there. – Cameron Buie Feb 27 '13 at 03:36
  • @CameronBuie *edit: Nvm I understand what you are trying to say now :). – Kot Feb 27 '13 at 03:47
  • @CameronBuie I am going over this problem again and got confused as to why the final answer is a negative when there aren't any negatives in the original function. The first step my professor took was dividing both the numerator and denominator by $-x^5$. What made him divide by a negative instead of a positive? – Kot Feb 28 '13 at 18:43
  • @StevenN: The big thing to remember is that for any positive even integer $n,$ we have $$\sqrt[n]{\alpha^n}=|\alpha|,\tag{#}$$ and for $\alpha<0$ we have $$|\alpha|=-\alpha.\tag{##}$$ He probably divided by $-x^5$ to get $-x^{-5}$ in the denominator. For $x<0,$ we can rewrite this as $|x^{-5}|$ by $(##),$ and then rewrite again as $$\sqrt{\left(x^{-5}\right)^2}=\sqrt{x^{-10}}$$ by $(#).$ Then we have $$\sqrt{x^{-10}}\sqrt{2x^{10}+x^2}=\sqrt{x^{-10}\left(2x^{10}+x^2\right)}=\sqrt{2+x^{-8}}$$ in the denominator, while his first step left $-(2+x^{-3})$ in the numerator. Same result. – Cameron Buie Feb 28 '13 at 18:57
  • @CameronBuie I am still confused as to why he divided by a negative instead of a positive. I did the same steps but divided by a positive and the answer came out as a positive $\sqrt{2}$. I think I am missing a concept here... – Kot Feb 28 '13 at 22:17
  • @StevenN: I have expanded my answer to hopefully give you a few more ways to see it. Let me know if you still have questions. – Cameron Buie Feb 28 '13 at 23:15
  • @CameronBuie I just watched a video by PatrickJMT and he explained it very clearly. Thank you for your answer though :D. – Kot Mar 01 '13 at 01:06
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Well, if you can't get rid of a square root, put everything in a square root. Notice that for $x$ sufficiently negative--in particular, for $x<-\frac1{\sqrt[3]2}$--we have $2x^5+x^2<0,$ so that $$2x^5+x^2=-|2x^5+x^2|=-\sqrt{(2x^5+x^2)^2}$$ for sufficiently negative $x$-values. Thus, for such $x$-values, we have

$$\begin{align} \frac{2x^5+x^2}{\sqrt{2x^{10}+x^2}} &= \frac{-\sqrt{(2x^5+x^2)^2}}{\sqrt{2x^{10}+x^2}} \\ &= -\sqrt{\frac{(2x^5+x^2)^2}{2x^{10}+x^2}} \\ &= -\sqrt{\frac{4x^{10}+4x^7+x^4}{2x^{10}+x^2}} \\ &= -\sqrt{2+\frac{4x^{10}+4x^7+x^4-2(2x^{10}+x^2)}{2x^{10}+x^2}} \\ &= -\sqrt{2+\frac{4x^7+x^4-2x^2}{2x^{10}+x^2}} \\ &= -\sqrt{2}\sqrt{1+\frac{4x^7+x^4-2x^2}{4x^{10}+2x^2}} \\ \end{align} $$

The fraction inside the right-hand square root is about $1/x^3$, so the right-hand square root goes to $1$ as $x\to-\infty$, so the whole thing goes to $-\sqrt{2}$.

Cameron Buie
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marty cohen
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