Prove the $\lim_{x \to -1} x^2 + 4x + 2 = 0$ is wrong by negating the $\epsilon-\delta$ definition of limit

Question

I'm struggling to disprove the following limit is wrong by negating the definition

$$ \lim_{x\to-1} x^2 + 4x + 2 = 0 $$

The negation of the definition applied to my case should be

$$ \exists \epsilon > 0 \forall \delta > 0 : 0 < |x+1| < \delta \text{ and } |x^2 + 4x + 2| \geq \epsilon $$

This is to me equivalent to find some $\epsilon > 0$ such that the following system has solution for each $\delta > 0$

$$ \left\{ \begin{array}{l} |x^2 + 4x + 2| \geq \epsilon \\ 0 < |x+1| < \delta \end{array} \right. $$

I started by picking $\epsilon = 1$, and assumed that $\delta \geq 1$ in such a case since $$ x^2 + 4x + 2 = (x + 1)^2 + 2(x + 1) - 1 $$

it follows that if $x = 0$ solves the system, I'm having trouble however when $0 < \delta < 1$. I cannot manage to construct a solution of that system... any suggestion?

Thank you

Answer 1

Take $\varepsilon=\frac34$. If $x\in\Bbb R$, then$$x^2+4x+2=(x+1)^2+2(x+1)-1.\tag1$$Therefore, for any $\delta>0$, if $-\delta<x+1<0$, and if $x+1\geqslant-\frac12$, you have $(x+1)^2\leqslant\frac14$ and $2(x+1)\leqslant0$. So, it follows from $(1)$ that $x^2+2x+2\leqslant-\frac34$, and therefore$$|x^2+4x+2|\geqslant\frac34=\varepsilon.$$

Answer 2

There is a missing quantifier: The definition of limit actually has an interior $\forall x$, so the negation should really be formalized as

$$\exists\epsilon\gt0\forall\delta\gt0\exists x\in\mathbb{R}:0\lt|x+1|\lt\delta\text{ and }|x^2+4x+2|\ge\epsilon$$

Added later (in response to comments from Rob Arthan and the OP): At this point, the issue becomes one of making a judicious choice of $\epsilon$ and then finding a "nice" relationship of $x$ to $\delta$ so that two inequalities are satisfied. The key idea is to keep $x$ away from either value for which $x^2+4x+2=0$ (i.e., $x=-2\pm\sqrt6$) even when $\delta$ is so large that the inequality $|x+1|\lt\delta$ allows it. There are many ways to do this, but one I like is to let $x=-1-{\delta\over1+\delta^2}$. It's clear that $|x+1|=\delta/(1+\delta^2)\lt\delta$ (when $\delta\gt0$), so the first inequality is satisfied. As for the second, we have

$$\begin{align} x^2+4x+2 &=(x+1)^2+4(x+1)-1\\ &={\delta^2\over(1+\delta^2)^2}-{4\delta\over1+\delta^2}-1\\ &={\delta^2-4\delta(1+\delta^2)-(1+\delta^2)^2\over(1+\delta^2)}\\ &=-{\delta^4+2\delta^3+\delta^2+2\delta+1\over\delta^4+2\delta^2+1} \end{align}$$

at which point it's easy to see, for example, that $|x^2+4x+2|\ge{1\over1000}$. (In fact, $\epsilon=1$ works, since showing that $\delta^4+2\delta^3+\delta^2+2\delta+1\ge\delta^4+2\delta^2+1$ for all $\delta\gt0$ reduces to verifying that $2\delta^2-\delta+2\ge0$, which is straightforward to check.)

Remark: The choice of $x=-1-\delta/(1+\delta^2)$ comes mainly from experience at finagling with limit problems of this type. Sometimes the first thing you try doesn't work, in which case you wad that scratch paper up and try try again. It can be fun to make choices that just barely work, but it's hardly imperative to do so.

Answer 3

Set $f(x) = x^2 + 4x + 2.$

There are actually two ways of proving that the stated limit is wrong. One way is the direct approach:

You can simply evaluate $f(x = -1)$ directly.

$f(-1) = (-1)^2 + 4(-1) + 2 = -1 \neq 0.$

Therefore the $\lim_{x \to -1} f(x) \neq 0.$

When $f(x)$ is well defined at the value that $x$ is going towards, (in this case, $[x = -1]$), as in this problem, this approach is viable.

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The alternative approach, which seems to be a requirement of the problem is to demonstrate that the limit is wrong through the negation of the $\epsilon,\delta$ approach.

Note: the comment of lone student makes a serious point. If you know that a function can't converge to two distinct points, then you could simply use the $\epsilon,\delta$ approach to show that $f(x)$ does in fact converge to $(-1)$ rather than $0$.

I will take the more pedantic approach:

Your negation has a flaw.

The approach should be:

$\exists ~\epsilon > 0~$ such that $\forall \delta > 0$ at least one value for $x$ can be found such that

$$0 < |x - (-1)| < \delta ~~\text{and}~~ |f(x) - 0| > \epsilon.\tag{Constraint-1}$$

I will simply set $\epsilon = (1/2).$

It is desired, given any specific value of $\delta > 0$, no matter how small $\delta$ is, to provide a formula for finding $x$ such that (Constraint-1) above is satisfied.

This means, that with $\epsilon$ fixed at $[\epsilon = (1/2)]$, you want to demonstrate that for any $\delta > 0$, some value of $x$ can be found such that:

$$-\delta < (x+1) < \delta ~~\text{and}~~ |(x^2 + 4x + 1)| > (1/2). \tag{Constraint-2}$$

Specify $x$ as follows: $x = \min[(\delta/2) -1, (1/10) - 1].$

This specification serves the following purposes:

• It guarantees that $x$ satisfies the LHS of Constraint-2.

• It is a general specification for $x$ in terms of $\delta$, as required.

• It guarantees that $-1 < x \leq -0.9$.

There are two possibilities, given how $\delta > 0$ is chosen.

If $(\delta/2) \leq (1/10)$, then $x = (\delta/2 - 1) \implies$

$f(x) ~=~ [\frac{\delta^2}{4} + 1 - \delta] + 4[(\delta/2) - 1] + 2$

$= ~ \frac{\delta^2}{4} + \delta - 1$.

Here, since it is assumed that $(\delta/2) \leq (1/10)$, you clearly have that $|f(x)| > (1/2)$, so the RHS of Constraint-2 is satisfied.

If $(\delta/2) > (1/10)$, then $x = (1/10 - 1) \implies$

$f(x) ~=~ [(1/100) + 1 - (2/10)] + 4[(1/10) - 1] + 2$

$= ~ (1/100) + (2/10) - 1$.

Again, you clearly have that $|f(x)| > (1/2)$, so the RHS of Constraint-2 is satisfied.

Answer 4

As indicated in my comment to question, let us take $\epsilon=1/2$. And let $\delta>0$ be given. We need to show that there is some $x$ with $0<|x+1|<\delta$ and $|x^2+4x+2|\geq 1/2$.

Let us observe that $$|x^2+4x+2|=|(x^2+4x+3)-1|\geq ||x+1||x+3|-1|$$ and the last expression is greater than or equal to $1/2$ if (not iff) $|x+1||x+3|<1/2$.

Let us assume that $|x+1|<1$ and then $|x+3|<3$ and the inequality $|x+1||x+3|<1/2$ is satisfied if $|x+1|<1/6$. Thus it should be clear that if $|x+1|<1/6$ then our job is done.

Thus we can take any value of $x$ which satisfies $0<|x+1|<\min(\delta, 1/6)$ and it will ensure that $|x^2+4x+2|\geq 1/2$.