Real analysis: prove the limit exists

[docnet]

Homework Statement

Prove the limit exists.

Relevant Equations

$\text{lim}_{x\rightarrow a}\quad f=L$ means $\forall \epsilon>0, \exists \quad \delta$ such that $|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$.

Prove that each of the limits exists or does not exist.

1. $\text{lim}_{x\rightarrow 2}(x^2-1)=3$

$\text{lim}_{x\rightarrow 2}(x^2-1)=3$ if $\forall \epsilon>0, \exists \delta$ such that $|x-2|<\delta \Rightarrow |f(x)-3|<\epsilon$.
\begin{align}&|x^2-1|=|x+1||x-1|\leq \epsilon\\
\Rightarrow &|x-1|\leq \frac{\epsilon}{|x+1|}\\
&\delta=\frac{\epsilon}{|x+1|}\end{align}

2. $\text{lim}_{x\rightarrow 1}(2x-1)\neq 2$
$\text{lim}_{x\rightarrow 1}(2x-1)\neq 2$ if the limit exists elsewhere than $2$. This means for all $\epsilon>0$, there exists a $\delta$ such that $|x-1|<\delta$ implies $\text{lim}_{x\rightarrow 1}(2x-1)= L$ for some $L\neq 2$.
\begin{align}&|2x-1-1|=2|x-1|\leq \epsilon\Rightarrow |x-1|\leq \frac{\epsilon}{2}\\
&\delta=\frac{\epsilon}{|2|}\end{align}

3. $\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})=1$
We use $x\rightarrow \frac{1}{x}$ to transform the limit.
\begin{align}&\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})\rightarrow \text{lim}_{x\rightarrow 0} (1+\frac{n}{n+1})=1\\
&|1+\frac{n}{n+1}-1| \leq |n| <\epsilon \Rightarrow |n|\leq \epsilon\\
&\delta=\epsilon\\
&\therefore \forall \epsilon>0, |n|<\epsilon \Rightarrow |1+\frac{n}{n+1}-1|<\epsilon\\
&\therefore \forall \epsilon>0, |n-\infty|<\epsilon \Rightarrow |1+\frac{1}{n+1}-1|<\epsilon\\
\end{align}

 

[Office_Shredder]

For number 1, your delta depends on x, which is generally not good practice. For number 2 I'm not even sure what you're trying to do. Are you trying to prove a limit exists? For number 3 there's a lot of confusion between x and n, which makes it hard to tell what's going on.I think it would help to take a step back and try to write the proof only in the "right" direction. For #1, start off by telling us what $\delta$ you want to pick for a given $\epsilon$, instead of writing down the steps you used to derive it (those steps can be useful for figuring out what you want the proof to look like, but they are not the actual proof typically).

 

[docnet]

For number 1, your delta depends on x, which is generally not good practice. For number 2 I'm not even sure what you're trying to do. Are you trying to prove a limit exists? For number 3 there's a lot of confusion between x and n, which makes it hard to tell what's going on.I think it would help to take a step back and try to write the proof only in the "right" direction. For #1, start off by telling us what $\delta$ you want to pick for a given $\epsilon$, instead of writing down the steps you used to derive it (those steps can be useful for figuring out what you want the proof to look like, but they are not the actual proof typically).

Okay, I understand.

 

[docnet]

1)

$\text{lim}_{x\rightarrow 2}(x^2-1)=3$ if $\forall \epsilon>0, \exists \delta$ such that $|x-2|<\delta \Rightarrow |f(x)-3|<\epsilon$.

Let $\delta<1$.

Then, $|x-2|<\delta<1\Rightarrow |x-2|<1 \Rightarrow -1<x-2<1$.

Adding 4 to the last inequality gives
$3<x+2<5\Rightarrow |x+2|<5$.

Then,
$|x^2-4|=|x+2||x-2|<5|x-2|<\epsilon\Rightarrow |x-2|<\epsilon/5$.

Select $\delta=\text{min}(1,\epsilon/5)$.

Check: $\forall \epsilon>0,\delta=\text{min}(1,\epsilon/5)$.
Then $|x^2-4|=|x+2||x-2|<5|x-2|<5\delta\leq 5\cdot \epsilon/5=\epsilon$.

 

[PeroK]

1)

$\text{lim}_{x\rightarrow 2}(x^2-1)=3$ if $\forall \epsilon>0, \exists \delta$ such that $|x-2|<\delta \Rightarrow |f(x)-3|<\epsilon$.

Let $\delta<1$.

Then, $|x-2|<\delta<1\Rightarrow |x-2|<1 \Rightarrow -1<x-2<1$.

Adding 4 to the last inequality gives
$3<x+2<5\Rightarrow |x+2|<5$.

Then,
$|x^2-4|=|x+2||x-2|<5|x-2|<\epsilon\Rightarrow |x-2|<\epsilon/5$.

Select $\delta=\text{min}(1,\epsilon/5)$.

Check: $\forall \epsilon>0,\delta=\text{min}(1,\epsilon/5)$.
Then $|x^2-4|=|x+2||x-2|<5|x-2|<5\delta\leq 5\cdot \epsilon/5=\epsilon$.

This is not bad. Although, you seem to have switched from $x^2 -1$ to $x^2 - 4$ at some point!

The proof proper should start with "Let $\epsilon > 0$". In any case, you introduce $\epsilon$ without saying what it is. At that point the proof gets a little confused between the proof itself and some supporting notes/calculations.

 

[docnet]

This is not bad. Although, you seem to have switched from $x^2 -1$ to $x^2 - 4$ at some point!

***sweats profusely***

Question: Does one always include the supporting notes (like the derivation of $\delta$) as part of the proof?3)

Supporting notes:
$$\displaystyle{\lim_{n \to \infty}}\Big(1+\frac{1}{n+1}\Big)=1\iff \displaystyle{\lim_{x \to 0}}\Big(1+\frac{x}{x+1}\Big)=1$$
Let $\delta<\frac{1}{2}$.

Then,
$$|x|<\delta<\frac{1}{2}\Longrightarrow |x|<\frac{1}{2}$$
$$\Longrightarrow |x+1|<\frac{3}{2}$$
$$\Longrightarrow \frac{x}{ |x+1|}<\frac{3|x|}{2}$$
Select $\delta = \text{min}(\frac{1}{2}, \frac{3\epsilon}{2})$.

The proof:

Let $\epsilon>0$ be given. $\displaystyle{\lim_{x \to 0}}\Big(1+\frac{x}{x+1}\Big)=1$ means there exists a $\delta$ for every $\epsilon>0$ such that
$$|x|<\delta\Longrightarrow |\frac{x}{x+1}|<\epsilon$$
Choose $\delta = \text{min}(\frac{1}{2}, \frac{3\epsilon}{2})$.
$$|\frac{x}{x+1}<\frac{2|x|}{3}$$
Since if $|x|< \frac{1}{2}, |x+1|<\frac{3}{2}$ and
$$\frac{2|x|}{3}<\frac{2\delta}{3}<\frac{2}{3}\cdot\frac{3\epsilon}{2}=\epsilon$$

 

[MidgetDwarf]

there's also somewhat of a short cut for this types of problems involving f(x)=x. Prove that the limit of f(x) exist at any point in R. Then prove that if two functions have a limit at a point c, then their sum and product also have a limit at that point. .

 

[Office_Shredder]

Docnet, I think the best way to start #1 is.

Let $\epsilon > 0$, and let $\delta =\min(0.9,\epsilon/6)$

(I picked 0.9 because your first step assumed $\delta$ was strictly less than 1! Similar for $\epsilon/6$ instead of 5)Then, because $\delta<1$, blah blah blah (eq. 1)

Since $\delta < \epsilon/5$, blah blah blah (eq 2).

Combining (1) and (2) yields [stuff]

3 is ok, but I wouldn't call it "supporting notes". Feel free to call it lemma #1, and then below say by lemma #1, we get the following. (Also note again your $\delta$ is slightly too large because of strict vs weak inequalities. If only Leibniz had defined continuity using weak inequalities the world would be a better place)

 

[PeroK]

@docnet I would be comfortable with the idea of intepreting the definition of a limit in terms of what needs to be shown in a particular case. For example:

(*) Let $\epsilon > 0$; choose $\delta ...$, then $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$.

Once you have that you can conclude that $\lim_{x \rightarrow a} f(x) = L$. That is enough.

You don't have to reconstruct a pseudo-definition of the limit with $\forall \epsilon > 0 ...$.

What's going on here might be an interesting question for the foundations of mathematics, but as far as real analysis is concerned it is enough to show (*).

 

[SammyS]

Prove that each of the limits exists or does not exist.

1. $\displaystyle \lim_{x\rightarrow 2}(x^2-1)=3$

2. $\displaystyle \lim_{x\rightarrow 1}(2x-1)\neq 2$

3. $\displaystyle \lim_{n\rightarrow \infty} (1+\frac{1}{n+1})=1$

I see that so far, this thread mostly addresses Problem 1.

I suggest that you post Problems 2 and 3 each in a thread of its own.

 

[docnet]

@docnet I would be comfortable with the idea of intepreting the definition of a limit in terms of what needs to be shown in a particular case. For example:

(*) Let $\epsilon > 0$; choose $\delta ...$, then $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$.

Once you have that you can conclude that $\lim_{x \rightarrow a} f(x) = L$. That is enough.

You don't have to reconstruct a pseudo-definition of the limit with $\forall \epsilon > 0 ...$.

What's going on here might be an interesting question for the foundations of mathematics, but as far as real analysis is concerned it is enough to show (*).

Thank you, I understand. Hopefully, I did better this time.

2)
Prove that $\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$.

Method 1:
Given $\epsilon>0$, choose $\delta=\epsilon/2$.
Then,
$$|x-1|<\delta\Longrightarrow |x-1|<\epsilon/2\Longrightarrow 2|x-1|<\epsilon$$
$$2|x-1|<\epsilon\Longrightarrow |2x-1-1|<\epsilon$$
So $$1=\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$$.

Method 2:
$$\neg \Big(\forall\epsilon>0, \exists \delta>0\quad \text{s.t.}\quad|x-1|<\delta\Longrightarrow |2x-1-2|<\epsilon \Big)$$
$$\equiv\exists \epsilon>0\quad\text{s.t.}\quad \forall \delta>0, \exists x\quad\text{s.t.}\quad (|x-1|<\delta)\land(|2x-1-2|\geq \epsilon)$$

Given $\delta>0$, choose $\epsilon=1$ and $x=1$.

$$|2x-1-2|\geq \epsilon\Longrightarrow |2\cdot 1-3|=|-1|=1$$
and
$$|x-1|<\delta \Longrightarrow |1-1|=0<\delta$$
By proving the negated statement, $\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$

 

[docnet]

there's also somewhat of a short cut for this types of problems involving f(x)=x. Prove that the limit of f(x) exist at any point in R. Then prove that if two functions have a limit at a point c, then their sum and product also have a limit at that point. .

This is truly useful information! Although, the professor told us we have to use the epsilon-delta definition of limit for these problems.

I see that so far, this thread mostly addresses Problem 1.

I suggest that you post Problems 2 and 3 each in a thread of its own.

After post #11, I hope that all three problems are almost all solved.

3 is ok, but I wouldn't call it "supporting notes". Feel free to call it lemma #1, and then below say by lemma #1, we get the following. (Also note again your $\delta$ is slightly too large because of strict vs weak inequalities. If only Leibniz had defined continuity using weak inequalities the world would be a better place)

Thank you, I understand. Why did Leibniz not define continuity using weak inequalities? (really curious).

 

[PeroK]

Thank you, I understand. Hopefully, I did better this time.

2)
Prove that $\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$.

Method 1:
Given $\epsilon>0$, choose $\delta=\epsilon/2$.
Then,
$$|x-1|<\delta\Longrightarrow |x-1|<\epsilon/2\Longrightarrow 2|x-1|<\epsilon$$
$$2|x-1|<\epsilon\Longrightarrow |2x-1-1|<\epsilon$$
So $$1=\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$$.

Method 2:
$$\neg \Big(\forall\epsilon>0, \exists \delta>0\quad \text{s.t.}\quad|x-1|<\delta\Longrightarrow |2x-1-2|<\epsilon \Big)$$
$$\equiv\exists \epsilon>0\quad\text{s.t.}\quad \forall \delta>0, \exists x\quad\text{s.t.}\quad (|x-1|<\delta)\land(|2x-1-2|\geq \epsilon)$$

Given $\delta>0$, choose $\epsilon=1$ and $x=1$.

$$|2x-1-2|\geq \epsilon\Longrightarrow |2\cdot 1-3|=|-1|=1$$
and
$$|x-1|<\delta \Longrightarrow |1-1|=0<\delta$$
By proving the negated statement, $\displaystyle{\lim_{x \to 1}}(2x-1)\neq 2$

Method 1 makes no sense to me at this time in the morning.

Method 2 is the right approach, but it's not right.

First, it should be $0 < |x - a| < \delta$ and that the $0 <$ is important. In other words, the limit is independent of $f(a)$. And, in fact, $f$ may be undefined at $a$. You cannot choose $x = 1$ when analysing the limit in this case.

Second, at the start of the proof I would say simply that you will show the definition does not hold for $\epsilon = 1$.

Third, at this point $\epsilon$ should disappear from your working, and be replaced by $1$. This is similar to the point I made above, where you seem reluctant to abandon the full generality of the definition.

So, the start of the proof should look like:

We show that the definition of the limit fails for $\epsilon = 1$. I.e. it is enough to show that:
$$\forall \delta > 0, \exists x: 0 < |x - 1| < \delta \ \text{and} \ |(2x -1) - 2| \ge 1$$That statement alone shows that a) you understand the negation of the limit definition; and b) shows specifically what you have to do in this case.

 

[WWGD]

A result you can use once you can figure these with a $\delta -\epsilon$ format is that if $f$ is continuous , then $[x \rightarrow x_0] \rightarrow [ f(x) \rightarrow f(x_0)]$.