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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?

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2  
$(2x^{10}+x^2)^{\frac12}$ is not $(2x^{10})^{\frac12}+x^{\frac12}$ –  ChairOTP Feb 27 '13 at 2:03
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I would recommend multiplying top & bottom by $x^{-5}$. –  Gerry Myerson Feb 27 '13 at 2:06
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This is way way too far from being correct. –  Kaster Feb 27 '13 at 2:06
    
I have multiplied the top and bottom by $x^{-5}$ but I am not sure what to do next. –  Kot Feb 27 '13 at 2:15
    
What happens to $\frac1x$ as $x$ approaches infinity? –  ChairOTP Feb 27 '13 at 2:20

2 Answers 2

up vote 5 down vote accepted

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, $$\begin{align}\lim_{x\to-\infty}f(x) &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2x^{10}}}\\ &= \lim_{x\to-\infty}\frac{2x^5}{\sqrt{2}\sqrt{x^{10}}}\\ &= \frac{2}{\sqrt{2}}\cdot\lim_{x\to-\infty}\frac{x^5}{\sqrt{(x^5)^2}}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}\frac{x^5}{|x^5|}\\ &= \sqrt{2}\cdot\lim_{x\to-\infty}-1\\ &= -\sqrt{2}.\end{align}$$

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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 2:39
    
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 2:42
    
@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 $x\to-\infty$ –  Babak S. Feb 27 '13 at 2:43
    
Nice, Cameron! +1 –  amWhy Feb 27 '13 at 2:45
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@StevenN: Ah! I see what happened. The next step would be to rewrite $$\cfrac{2x^5+x^2}{-x^5\sqrt{2+\frac1{x^8}}}=\cfrac{2+\frac1{x^3}}{-\sqrt{2+\fra‌​c{1}{x^8}}},$$ 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 3:36

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 $-\sqrt{2}$.

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That doesn't work when the numerator is negative, Marty, but you're close. –  Cameron Buie Feb 27 '13 at 2:38
    
Good point. But my formatting is pretty good:) –  marty cohen Feb 28 '13 at 2:19
    
Indeed. All you really need to do to fix it is to say that for $x$ sufficiently negative, we have $$2x^5+x^2=-\sqrt{\left(2x^5+x^2\right)^2},$$ and so all the rest is basically right, just missing a $-$ sign. –  Cameron Buie Feb 28 '13 at 2:51

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