Why is $\varepsilon x^5 \sim -x$? I'm trying to understand what's going on in this lecture on perturbation (the link brings you to 1h 08m 12s). 
The original problem is to find the real root of $$x^5+x=1.$$ We have inserted $\varepsilon$ as follows $$\varepsilon x^5+x=1$$ and are discussing which of the three terms ($\varepsilon x^5$ or $x$ or $1$) is negligible as $\varepsilon \to 0$ so we could throw it way.
So we consider all three cases and it turns out that:


*

*Term $1$ is negligible and we learn something if we remove it,

*Term $x$ is not negligible,

*Term $\varepsilon x^5$ is negligible but we learn nothing if we remove it.
My question is about term $1$ being negligible. On the blackboard, we have $$\varepsilon x^5 \sim -x \quad \text{as} \quad \varepsilon \to 0$$ and claim that it's true. Why?
I thought it means $$\lim_{\varepsilon \to 0}\frac{\varepsilon x^5}{-x} = 1.$$ In fact, this is how it was defined it at the beginning of the lecture.
So what's going on here?
 A: I think the best way to look at this issue is to write the equation as
\begin{equation}
\epsilon x^5 = 1-x. \tag{1}
\end{equation}
When $\epsilon=1$, the situation is as shown in the picture below. Since $x^5$ is monotonically increasing (and $1-x$ is strictly monotonically decreasing), it's clear that $(1)$ has one real solution, where the two curves intersect. 
Now, what happens if we take $\epsilon$ smaller and smaller? Then, the curve $\epsilon x^5$ will be 'flatter and flatter', until for $\epsilon = 0$, the curve coincides with the horizontal axis. From the figure, it is very clear what will happen to the point of intersection: as $\epsilon \to 0$, this point will converge to $x=1$, from the left.

However, in the lecture, the relation $\epsilon x^5 \sim - x$ is not written down in the context of this only real root! The real root, which is close to $1$, is found on the right side of the blackboard. What the lecturer tries to accomplish by looking at the asymptotic regime where $\epsilon x^5 \sim - x$, is to find out if we can find an approximate expression for the other four roots, which are all fully complex. As he demonstrates, these can be approximated as
\begin{equation}
 x \sim \pm\frac{1\pm i}{\sqrt{2}}\frac{1}{\epsilon^{\frac{1}{4}}}.
\end{equation}
You can see that for these complex roots, we indeed have
\begin{equation}
 \lim_{\epsilon \to 0} \frac{\epsilon x^5}{-x} = \lim_{\epsilon \to 0} -\epsilon x^4 = \lim_{\epsilon \to 0}  - \left(\frac{1\pm i}{\sqrt{2}}\right)^4 = 1
\end{equation}
The idea behind the asymptotic statement '$\epsilon x^5 \sim - x$', as the lecturer explains in the video, is that we're looking for solutions to $(1)$ where $x$ is very large. In particular, we assume that $x$ is sufficiently large such that both $x \gg 1$ and $\epsilon x^5 \gg 1$. That means that these two terms dominate the equation, and we can neglect the term '$1$'. Another way to look at this is to divide $(1)$ by $x$, yielding
\begin{equation}
 \epsilon x^4 = \frac{1}{x} - 1. \tag{2}
\end{equation}
You can see that when $x$ is large, then the term $\frac{1}{x}$ goes to zero, so it is definitely much smaller than the term $-1$, and if you take $x$ large enough, then $\frac{1}{x}$ can also be made much smaller than $\epsilon x^4$. It is precisely this asymptotic regime that the lecturer is treating.
