$\sqrt{1\pm10\varepsilon+\varepsilon^2}=1\pm P(\varepsilon)$. Is there a better way than mine to find $P(\varepsilon)$?

Question

Some days ago we did a classwork, and there was this exercise:

Using the limit definition, verify $$\displaystyle \lim_{x\to0} \frac{3x^2-1}{x+1}=-1.$$

From $\displaystyle \left\lvert\frac{3x^2+x}{x+1}\right\rvert < \varepsilon$ we get the system $$\displaystyle \left\{ \begin{array}{c} \frac{3x^2+(1+\varepsilon)x+\varepsilon}{x+1}>0 \\ \frac{3x^2+(1-\varepsilon)x-\varepsilon}{x+1}<0 \\ \end{array} \right. $$ In search of the solution set, we study the sign of the two numerators, seeing when they are positive. This turns out respectively in the following intervals: $$\displaystyle x<\frac{-1-\varepsilon-\sqrt{1-10\varepsilon+\varepsilon^2}}{6} \vee x>\frac{-1-\varepsilon+\sqrt{1-10\varepsilon+\varepsilon^2}}{6} \\ x<\frac{-1+\varepsilon-\sqrt{1+10\varepsilon+\varepsilon^2}}{6} \vee x>\frac{-1+\varepsilon+\sqrt{1+10\varepsilon+\varepsilon^2}}{6}.$$

Now, my professor suggested we cancelled out $\varepsilon^2 \pm 10\varepsilon$ from the radicands, hence easily getting to the solution of the exercise. But to me, this is cheating. I mean, if we cancel out that, then how about the other $\varepsilon$'s that appear? If we wanted to be coherent, we would have to cancel out those as well, but then the whole verification makes no sense.

Thus, I want to find a legit way to do it. I have tried many other things, but always finding stuff that involves $1$ again, usually under a root. So far my best try is this:

let $\varepsilon=10^{-m}$, $m$ arbitrarily large. Then we have, recalling the first two inequations, $$\displaystyle \sqrt{1\pm10^{-m+1}+10^{-2m}}=\sqrt{\frac{1+10^{m+1}\left(10^{m-1}\pm1\right)}{10^{2m}}}=\frac{\sqrt{1+10^{m+1}\left(10^{m-1}\pm1\right)}}{10^m}.$$ Numerically I have found that the numerator is very nicely approximated, with $m>2$, by $10^m \pm 5 - 10^{-m+2}$, so the whole fraction is $\approx 1\pm5\times10^{-m}-10^{-2m-2}=1\pm5\varepsilon-100\varepsilon^2.$ Since $m$ is arbitrarily large, the digits of the value of the root that we're not taking into account become kind of truly insignificant (I mean, they would appear after arbitrarily many $9$'s, so unless we get into the hyperreals, I'd think they actually are wiped out, but I might be wrong). Hence this allows us (I think) to write $$ \sqrt{1\pm10\varepsilon+\varepsilon^2}=1\pm5\varepsilon-100\varepsilon^2.$$ If anything, I've found a good approximation. But in this case I would not be satisfied at all with my result.

Edit: Alright, it wasn't necessary to do all this. But is mine an approximation or not?

Answer 1

Let $\delta<1$ and assume $|x|<\delta.$ Then

$$\displaystyle \left\lvert\frac{3x^2+x}{x+1}\right\rvert =\frac{|x||1+3x|}{|x+1|}\le \frac{\delta (1+3\delta)}{1-\delta}=\frac{\delta +3\delta^2}{1-\delta}\le \frac{4\delta}{1-\delta}. $$

Now,

$$\frac{4\delta}{1-\delta}<\epsilon \Leftrightarrow \delta<\frac{\epsilon}{4+\epsilon}.$$

Answer 2

You are overcomplicating things. Note that, if $x>0$

$$\tag{1} \left\lvert\frac{3x^2+x}{x+1}\right\rvert=|x|\left\lvert\frac{3x+1}{x+1}\right\rvert\leqslant 3|x|$$

In fact, solving $(3x+1)(1+x)=-3$ gives the solution $-2/3$. Thus, for $x\in (-2/3,\infty)$, $(1)$ holds. You can conclude that taking $\varepsilon=\delta/3$ in such domain works.