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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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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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Well, if you can't get rid of a square root, put everything in a square root. $\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 square root is about $1/x^3$, so the square root goes to 1, so the whole thing goes to $\sqrt{2}$. |
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