Prove that $\lim_{x\to 1}\left(4+x-3x^{3}\right)=2$.

I was inspired by this <link> on page 10-11, where it was shown that $$\lim_{x\to 1}\left(4+x-3x^{3}\right)=2.$$ I wonder if it's possible to show this way without choosing a good value of $\delta>0$ (for example $\delta\leq 1$ in the link). Since $$\left | (4 + x-3x^{3})-2 \right |=\left | x-1 \right |\left | -3x^{2}-3x-2 \right |,$$ and if $\left | x-1 \right |<\delta $ then we have two inequalities $$3(-\delta-1)<-3x<3(\delta-1)$$ and $$-3(\delta+1)^{2}<-3x^{2}<-3(-\delta+1)^{2}.$$ Altogether we get $$-3(\delta+1)^{2}+3(-\delta-1)-2<-3x^{2}-3x-2<-3(-\delta+1)^{2}+3(\delta-1)-2,$$ that is, $$-9\delta-8-3\delta^{2}<-3x^{2}-3x-2<9\delta-8-3\delta^{2}.$$ Is it equivalent to $$\left | -3x^{2}-3x-2 \right |<9\delta+\left | -8-3\delta^{2} \right |=9\delta+8+3\delta^{2}?$$ If yes, then we finally have $$\left | x-1 \right |\left | -3x^{2}-3x-2 \right |<\delta(9\delta+8+3\delta^{2})$$ Now the question is: is there no way to isolate or solve the positive value of $\delta$ in the inequality, $$\delta(9\delta+8+3\delta^{2})<\epsilon?$$ If the solution is found, then I would prove this way:

Let $\epsilon>0$ be given. Choose $\delta>0$ such that $\delta <\textrm{something}$. If $0<\left | x-1 \right |<\delta$ then $$\left | (4 + x-3x^{3})-2 \right |=\left | x-1 \right |\left | -3x^{2}-3x-2 \right |<\dots<\epsilon$$ for all $x$ in the domain.

Solution 1:

Its a great tragedy that we have to find an expression for $\delta$ in terms of $\epsilon$ to establish certain limits. I think most of crap textbooks are to be blamed for it.

If we carefully note the definition of limits then we see that we only need to have the existence of a certain $\delta > 0$ for any pre-assigned $\epsilon > 0$ such that some well defined inequalities hold. It is nowhere mentioned that we need to have $\delta$ expressed in terms of $\epsilon$ by means of a formula.

To establish that $\lim_{x \to 1}(4 + x - 3x^{3}) = 2$ we need to analyze the difference $$|(4 + x - 3x^{3}) - 2| = |x - 1||3x^{2} + 3x + 2|$$ Also we need to analyze this expression when $x$ is near $1$. So we can already fix $|x - 1| < 1$ i.e $0 < x < 2$. This would mean that $$2 < 3x^{2} + 3x + 2 < 20$$ Thus we can see that $$|(4 + x - 3x^{3}) - 2| < 20|x - 1|$$ when $0 < |x - 1| < 1$. Now we have to make the expression $|(4 + x - 3x^{3}) - 2|$ less than a pre-assigned positive number $\epsilon$ and it can be done if $20|x - 1| < \epsilon$. It is now obvious that we need to choose $\delta$ such that $0 < \delta < \min(1, \epsilon/20)$ and this will ensure that whenever $0 < |x - 1| < \delta$ then we have $$|(4 + x - 3x^{3}) - 2| < \epsilon$$ There is no need to find an exact value of $\delta$ in terms of $\epsilon$.

Another common mistake which students do (also visible in the approach of OP) is that they try to start with $0 < |x - a| < \delta$ and try to do algebraic manipulation to reach $|f(x) - L| < g(\delta)$ and then try to solve the equation $g(\delta) = \epsilon$.

This is so wrong an approach. You cannot convert "Analysis" into "Algebra" just like that. You need to analyze the expression $|f(x) - L|$ properly (like do some algebraic manipulation to simplify it) so that its dependence on the difference $|x - a|$ is visible very clearly.

In doing this analysis you may want to already limit the values of $x$ by choosing some bound for $|x - a|$ (like I chose $|x - 1| < 1$) and probably this restriction of values of $x$ will help to analyze the difference $|f(x) - L|$ even more clearly. A good goal is establish an inequality of type $$|f(x) - L| < K|x - a|$$ whenever $|x - a| < c$. Here $K$ and $c$ should be actual numerical expressions like $2, 3, \sqrt{2}, 2 + \sqrt{3}$. The value of $c$ is choosen suitably based on the problem and the value of $K$ comes out if we use the assumption $|x - a| < c$ suitably. Now it is easy to see that any $\delta$ with $0 < \delta < \min(c, \epsilon/K)$ does the job.

Solution 2:

There is a flaw in your argument that is worth noting. You write "if $|x−1|<\delta$ then we have two inequalities $$3(−δ−1)<−3x<3(δ−1)$$ and

$$−3(δ+1)^2<−3x^2<−3(−δ+1)^2."$$

The first of these is correct, but the second is only correct if you've already assumed $\delta\le1$. For example, one way to satisfy $|x-1|\lt\delta$ is to let $x=0$ and $\delta=2$. This turns the second line of inequalities into

$$-27\lt0\lt-3$$

Solution 3:

Say $f(x) = 4 + x - 3 x^3$. Now $f'(x) = 1 - 9 x^2$, $\ f'(x) < 0 $ on $(1/3, \infty)$ so $f$ maps $( 1/3, \infty)$ bijectively onto $( -\infty, 38/9)$. We have $\delta \mapsto 1+\delta \mapsto f(1+\delta) = 1+ \epsilon \mapsto \epsilon$ for $|\delta | < 2/3$ and we want the inverse function. Well, $\delta \mapsto \epsilon = f(1+\delta) - 2 = - 8 \delta - 9 \delta^2 - 3 d^3$. The inverse function $ \epsilon \mapsto \delta$ is also analytic and has a series expansion \begin{eqnarray} \delta = -\frac{\epsilon }{8}-\frac{9 \epsilon ^2}{512}-\frac{69 \epsilon ^3}{16384}-\frac{2565 \epsilon ^4}{2097152}+O[\epsilon ]^{5} \end{eqnarray} $\epsilon$ is a series in $\delta$, obtained by inverting the series ( polynomial in fact) $- 8 \delta - 9 \delta^2 - 3 d^3$. This series is convergent for $|\epsilon| <f(1/3) -2 = 38/9 -2 = 20/9= 2.222...$

Solution 4:

Here is a solution along the lines you are suggesting:

Let $\epsilon>0$ be given, and choose $\delta>0$ with $\delta<\min\{\sqrt[3]{\frac{\epsilon}{9}}, \sqrt{\frac{\epsilon}{27}}, \frac{\epsilon}{24}\}$.

If $0<|x-1|<\delta$, then $\big|(4+x-3x^3)-2\big|=\big|2+x-3x^3\big|=\big|(x-1)(-3x^2-3x-2)\big|$

$=\big|x-1\big|\big|3x^2+3x+2\big|=\big|x-1\big|\big|3(x-1)^2+9(x-1)+8\big|\le\big|x-1\big|\big(3|x-1|^2+9|x-1|+8\big)$

$\;\;\;<\delta(3\delta^2+9\delta+8)=3\delta^3+9\delta^2+8\delta<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$.