# How to disprove this limit?

1. Jun 15, 2013

### Alpharup

Well've I recently passed my school and am entering college...In my school, I have been exposed to intuitive calculus...ie...I learnt only the basic idea of limits, continuity ad differentiability...Then I proceeded to integral calculus, applications of calculus, differential equations, etc....After learning all these, I found that I was in a level for learning the exact definition of limit, continuity, etc...

Now, I browsed through a lot of websites for learning the epsilon-delta definition of limit..Though I can get an idea that the intuitive treatment of limit exactly equals the formal definition, I'am not satisfied..If I look in one angle, I find that I understood the definition..If I approach in another angle, I find that I do not understand anything about it...This is because I want to prove that a certain limit is wrong and Iam not able to prove by this definition....
For example...
we have to prove

lim 2x+1=5...
x→2
I will give the method of solving this problem...Please point me if my method is wrong..

The actual definition of limit is

Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. Then the statement

lim f(x)=L
x→c
means

for all real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c | < δ, we have |f(x) − L| < ε ( From Wikipedia)

1. By the definition of limit, I assume that there exists a δ > 0 such that for all x with 0 < |x − c | < δ..

2. Then I also assume that |f(x) − L| < ε is true..
3. Then by using these assumptions, I prove that ε exists and is >0....which is accoardance with the above definition..
4.Since all the conditions satisfy, the limit is true...

Now, I give this contradictiory statement

lim 2x+1= 5.1
x→2

How do you disprove this limit by the formal definition? If a certain statement which is true can be proved by a definition , then a certain statement which is false can also be disproved by the same definition...

I just want to disprove the statement, please help me out with this concept....

2. Jun 15, 2013

### HallsofIvy

As you say, if $$\lim_{x\to a} f(x)= L$$, then, by the definition of limit, "given any $\epsilon> 0$ there must exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon[/tex]". To show that a purported limit is NOT correct you must show that this is NOT true- that there exist some $$\epsilon$$ such that such a delta does not exist. Of course we know that [itex]\lim_{x\to 2} 2x+ 1$ is NOT 5.1 because we know the limit is actually 5. So choose $$\epsilon$$ smaller than the difference. For example, we can take $\epsilon= .05$
If $|x- 2|< \delta$, that is if $-\delta< x- 2< \delta$ or $2-\delta< x< 2+ \delta$, then $4- 2\delta< 2x< 4+2\delta$ and $5- 2\delta< 2x+ 1< 5+ 2\delta$. We just need to choose $\delta$ small enough that $$5+ 2\delta$$ is still smaller than 5.1- .05t= 5.05.

3. Jun 15, 2013

### lurflurf

In general to disprove
lim f=L
we would show that for some ε there does not exist δ such that
|f-L|<ε whenever |x-a|<δ

4. Jun 19, 2013

### Alpharup

I cant't undertstand this statement. Please explain me once again..

5. Jun 19, 2013

### Stephen Tashi

There's a typo. I think it should say "smaller than 5.1 - .05 = 5.05".

That is a suggestion about how to show that for $\epsilon = .05$ and for any $\delta > 0$ there exists an $x$ in $(2-\delta,2+\delta)$ and $x \ne 2$ such that $f(x)$ is not in $(5.1 - .05, 5.1 + .05)$, which is one way of saying what needs to be proven.

How to reformulate a "negation" of a statement containing quantifiers such as one of the form "It is not true that ( for each....there exists...for each....)" can be done systematically using the rules of logic that apply to quantifiers like "for each". Some people learn to do this "by common sense" without studying the formal rules, but most learn faster with some formal instruction.

A Proof:

Let $\epsilon = .05$. For any positive number $\delta$, the interval $(2-\delta, 2+\delta)$ contains a number $x$ such that $x \ne 2$ and $x$ is also in $(2 - \frac{.05}{2}, 2 + \frac{.05}{2}$. That is because the intervals $(2-\delta, 2+\delta)$ and $(2 - \frac{.05}{2}, 2 + \frac{.05}{2})$ have a non-empty intersection. (Regardless of which interval is larger, they must have some overlap besides the value 2 since they both contain numbers close to the number 2.)

Select $x$ to be a number such that $x \ne 2$ and $x$ is in both intervals. Then since $2 - \frac{.05}{2} < x < 2 + \frac{.05}{2}$ it follows that $4 - .05 < 2x < 4 + .05$ and $5 - .05 < 2x + 1 < 5 + .05)$ This shows $f(x)$ (which is $2x+1$ ) is not in $( 5.1 - 0.5, 5.1 + 0.5)$ because the intervals $(5.0 - .05, 5 + .05)$ and $(5.1 -.05, 5.1 + .05)$ have an empty intersection.

From the viewpoint of human understanding, the problem with the above proof is that numbers like $\epsilon = .05$ and $\frac{.05}{2}$ appear to be pulled out of the air. HallsOfIvey is suggesting a method of "working backwards" to figure out how to pick numbers that work.

There is no single standard for writing a correct proof. Some teachers and graders might accept the process of working backwards as the proof without making you write a more formal "forwards" proof. (It's like the situation in proving trigonometric identities. Many teachers accept writing down the identity to be proven and working backwards to some equality that's true to be a proof. )

Some teachers might not like the above proof's statements about intervals having non-empty intersection and they would want to see things phrased in terms of expressions that use absolute value signs.

By practice, you can learn to somewhat mechanically do such proofs for linear functions like f(x) = 2x + 1. Proofs of more complicated functions require a virtuosity with reasoning about inequalities that goes beyond mere mechanical manipulations of "doing the same thing to both sides". However, most introductory calculus courses don't expect you to do proofs with more complicated functions.

Last edited: Jun 19, 2013