Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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? –  Emmad Kareem 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
add comment

5 Answers

up vote 4 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
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
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $\lim\limits_{x \to 2} x + 3 \neq 6$.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.