# Prove that the limit does not go to 6

I want to prove that

$$\lim_{x \to 2} \ x + 3 \ne 6$$

What I thought about doing was first assuming the limit actually equaled $6$. Then taking an $x$ below and above $3$ and then finding a contradiction form the two statements

1) choosing $x = 1.5$ we get $|0.5|< \delta \implies |0.5| < \epsilon$

2) choosing $x = 2.5$ we get $|0.5|< \delta \implies |1.5| < \epsilon$

Then choose an epsilon equal $1$ and use the contrapostive of $2$ but I am not sure exactly how to phrase this.

• If you're allowed to assume uniqueness of limits, you could prove that the limit is 5 (which is very easy if you're allowed to assume continuity of polynomial functions) – Daniel Freedman Dec 29 '11 at 20:51
• since f(2) exists, then the limit is f(2) = 5 - What is the problem here? – NoChance Dec 29 '11 at 21:07
• Deriving such basic results without letting us know which theorems and definitions you want to allow is almost impossible for us readers. – Listing Dec 29 '11 at 21:09
• Is it not easy so see that the limit is 5? What other method do you want to use? – smanoos Dec 29 '11 at 21:09

Here is the definition of limit:

$\lim\limits_{x\rightarrow a} f(x)=L$ if for every $\epsilon>0$, there is a $\delta>0$ such that $$|f(x)-L|<\epsilon\quad\text{whenever }\quad 0<|x-a|<\delta.$$

What does it mean that $\lim\limits_{x\rightarrow a} f(x)\ne L$?

Well, it means (imprecisely) that there is an $\epsilon>0$ such that $f(x)$ is not close to $L$ no matter how close $x$ is to $a$.

More precisely it means that there is an $\epsilon>0$, such that no matter how small $\delta>0$ is, there is an $x$ with $0<|x-a|<\delta$ and yet $|f(x)-L|\ge\epsilon$.

So, in your case, you need to find a fixed value of $\epsilon$ such that for any $\delta>0$ there is an $x$ such that the following holds: $$\tag{1}|(x+3)-6|\ge\epsilon\quad\text{and}\quad 0<|x-2|<\delta.$$

Here, you could choose $\epsilon=1/2$.

Given any $\delta>0$, choose any $x$ such that $0<|x-2|<\min\{\delta,1/2\}$.

Then $x$ would be in the interval $(1.5,2.5)\,$. Consequently, $x+3$ would be in the interval $(4.5,5.5)$ and thus at least $1/2$ units away from 6. That is, $|(x+3)-6|\ge1/2$.

Informally, if $x$ is very close to 2, then $x+3$ would be far away from 6. And so there would be no $\delta$ that "works" in the definition of limit. The quantity $x+3$ is at least 1/2 unit away from 6 whenever $x$ is within 1/2 of 2.

Write down the formal definition of limit: $\lim_{x\to 2} x+3 = 6$ means: $$(\forall \epsilon>0)(\exists \delta>0)(\forall x)\Bigl[0<|x-2|<\delta \Rightarrow |(x+3)-6|<\epsilon\bigr]$$ Since you want to prove that this is not true, negate the statement: $$\neg(\forall \epsilon>0)(\exists \delta>0)(\forall x)\Bigl[0<|x-2|<\delta \Rightarrow |(x+3)-6|<\epsilon\bigr]$$ Push the negation through the quantifiers to get $$(\exists \epsilon>0)(\forall \delta>0)(\exists x)\Bigl[0<|x-2|<\delta \not\Rightarrow |(x+3)-6|<\epsilon\bigr]$$ Negating $P\Rightarrow Q$ yields $P\land \neg Q$, so the property to prove is $$(\exists \epsilon>0)(\forall \delta>0)(\exists x)\Bigl[0<|x-2|<\delta \land |(x+3)-6|\ge\epsilon\bigr]$$ In other words, we need to find some $\epsilon$ such that for all $\delta$ there is an $x$ closer to $2$ than $\delta$ such that $|x+3-6|=|x-3|$ is larger than $\epsilon$. This $x$ is allowed to depend on $\delta$, but we must find an $\epsilon$ that works for every $\delta$.

Thus, it is wrong when in your argument you start by setting $x=1.5$ and $x=2.5$ without speaking of $\epsilon$ and $\delta$ first. Neither of these $x$'s can possibly work for $\delta=0.001$, for example.

Hint: $\epsilon = \frac 12$ works. Can you see why?

Note that when we negate the definition, the "burden of proof" reverses. When we want to show what the limit is, the adversary chooses an $\epsilon$, and we must then find a $\delta$ that works for every $x$ that the adversary picks afterwards. But when we want to show what the limit is not, we get to pick $\epsilon$ and (later) $x$, whereas the adversary tries to find a $\delta$ that will foil us.

• Tiny nitpick: The value of the function at $2$ does not matter for the limit; so the correct definition of the limit is: for all $\epsilon > 0$, there exists $\delta >0$ such that $0 < |x - 2|<\delta \implies |x+3-6|<\epsilon$. – Srivatsan Dec 29 '11 at 21:18
• Ha, beat you by 2 seconds! – Henning Makholm Dec 29 '11 at 21:19
• Right. Henning 1, Srivatsan 0. :-) – Srivatsan Dec 29 '11 at 21:20

If the limit is $6$ and $\varepsilon=1/2$, then there exists $\delta>0$ such that whenever $2-\delta<x<2+\delta$ and $x$ is not exactly $2$, then $6-\varepsilon<x+3<6+\varepsilon$. That would mean $2.5<x<3.5$ whenever $2-\delta<x<2+\delta$ and $x\ne2$. No matter what positive number $\delta$ is, there will be some numbers in $(2-\delta,2+\delta)$ that are not in $(2.5,3.5)$. There you have a contradiction.

Let $L$ be some number other than $5$. Let's define $d=|L-5|$, and because $L\neq 5$ we have $d>0$.

The reverse triangle inequality says that for any $a$ and $b$, $$|a-b|\geq ||a|-|b||.$$ In particular, $$\left|L-(x+3)\right|=\left|(L-5)-\left(x-2\right)\right|\geq |d-|x-2||$$

Suppose that $$\lim_{x\to 2}\;(x+3)=L,$$ i.e. for any $\epsilon>0$, there exists an $\delta>0$ such that: for all $x$ with $|x-2|<\delta$, $$\left|L-(x+3)\right|<\epsilon.$$ Then we have that for all $x$ with $|x-2|<\delta$, $$|d-|x-2||<\epsilon$$ since $|d-|x-2||\leq|L-(x+3)|$.

But $x=2$ will always have $|x-2|=0<\delta$ for any $\delta>0$, so in particular we must have $$|d-|2-2||=|d|=d\leq\epsilon$$ for all $\epsilon>0$. But this is false, e.g. take $\epsilon=\frac{d}{2}$.

Thus, the limit cannot be any number other than 5, so the limit certainly cannot be 6.

The statement $\lim\limits_{x \to 2} x + 3 = 6$ is equivalent to saying that for any $\epsilon>0$ we have some $\delta>0$ such that $0<|x - 2|<\delta\implies |x+3-6|<\epsilon$. In order to prove $\lim\limits_{x \to 2} x + 3 \neq 6$, we want to find some $\epsilon>0$ such that this is not true. Let's try $\epsilon = 1/2$. What we want is to find points $x_1,x_2,\ldots$ which are arbitrarily close to $2$ (so that we have some point $x_n$ such that $0<|x_n-2|<\delta$ for any $\delta>0$) such that $|x_n+3-6|>1/2$ for all $n$. How about $x_n = 2 +\frac{1}{n+2}$? Well, $|x_n - 2| = \frac{1}{2+n}$, so the points get arbitrarily close to $2$, and $|x_n+3-6| = 1-\frac{1}{n+2}>1/2$ for all $n$, so $\lim\limits_{x \to 2} x + 3 \neq 6$.

• Actually, the value of the function at $2$ does not matter for the limit; so the correct definition of the limit is: for all $\epsilon > 0$, there exists $\delta >0$ such that $0 < |x - 2|<\delta \implies |x+3-6|<\epsilon$. [That said, the meat of the answer does not change due to this.] – Srivatsan Dec 29 '11 at 21:17
• @Srivatsan Thanks for the correction. – Alex Becker Dec 29 '11 at 21:30

There are seven steps to be followed when proving that the limit of f(x) as x->a is not = L, with the addition to negation shown above by David Mitra. 1. Find the real limit M for f(x) 2. Choose epsilon so that real limit M is not in (L-eps., L+ eps.) 3. Write the definition in a negated form 4. Remove absolute value for f(x)-L >/= eps. , and find x intervals using the choosen eps. 5. Observe the interval where a falls in for delta>0, that will mean that the intersection of that interval with (a-delta, a+delta) is not = empty set. 6. So whatever delta is,let x' be any point within an intersection. 7. Then absolute value of f(x')-L >/=eps. as required. Hence lim f(x) as x->a is not =L