Continuity mk3

• Mar 19th 2009, 08:27 AM
Showcase_22
Continuity mk3
Quote:

Prove that $f(x)=\begin{cases} x^2 \ \ \ \ \ \ \ \ \ x<0\\ sin(x) \ \ \ \ x \geq0 \end{cases}$ is continuous at the origin.
Sorry for posting so many of these! I'm finding them a little tricky and keep running into problems (Angry)

I've done two proofs for this, one uses $\epsilon$ and $\delta$, the other uses sequences.

The epsilon-delta version:

I need to find $|f(x)-f(x_0)|<\epsilon$ when $0<|x-x_0|< \delta$.
Since i'm trying to find continuity at 0, let $x_0=0$.

Therefore I need to find $|f(x)|< \epsilon$ when $0<|x|< \delta$.

For $x>0$, $|f(x)|< \epsilon \Leftrightarrow \ |sin(x)|< \epsilon \Leftrightarrow \ |x|< \epsilon$. Hence pick $\delta=\epsilon$.

For $x<0$, $|f(x)|<\epsilon \Leftrightarrow \ |x^2|< \epsilon \Leftrightarrow \ |x|< +\sqrt{\epsilon}$. Hence pick $\delta=+\sqrt{\epsilon}$

Therefore choose $\delta=min \{\epsilon, +\sqrt{\epsilon} \}$
$\color{red}{Is \ this \ the \ correct \ way \ to \ write \ \delta?}$

That was preliminary work so I have to write out the proper proof:

Let $\epsilon>0$ and define $\delta:=min \{ \epsilon, +\sqrt{\epsilon} \}$. When $0<|x|< \delta$:

$|f(x)-f(x_0)|=|x^2|< (\sqrt{\epsilon})^2=\epsilon$ for $x<0$.
$|f(x)-f(x_0)|=|sin(x)| \leq|x|< \epsilon$ for $x>0$.
Hence f is continuous $\forall x \in \mathbb{R}$.

__________________________________________________ ____________

Now for the limits version.

Let $a_n=-\frac{1}{n}$ and $b_n=\frac{1}{n}$.

Clearly $\lim_{n \rightarrow \infty}a_n=\lim_{a_n \rightarrow 0-} f(a_n)$ and $\lim_{n \rightarrow \infty}b_n=\lim_{b_n \rightarrow 0+} f(b_n)$.

I need to show that $\lim_{a_n \rightarrow 0-}f(a_n)=\lim_{b_n \rightarrow 0+}f(b_n)$ if $\lim_{x \rightarrow 0}f(x)$ exists.

$f(\lim_{n\rightarrow \infty}a_n)=f(\lim_{n \rightarrow \infty}\frac{1}{n})=0$
$f(\lim_{n\rightarrow \infty}b_n)=f(\lim_{n \rightarrow \infty}-\frac{1}{n})=0$

Therefore $\lim_{a_n \rightarrow 0-}f(a_n)=\lim_{b_n \rightarrow 0_+}f(b_n)$ so $\lim_{x \rightarrow 0}f(x)$ exists.

Hence the function is continuous at the origin.

$\color{red}{Is \ this \ correct?}$

__________________________________________________ ________

Thanks in advance to anyone who posts! (Rofl)
• Mar 19th 2009, 10:10 AM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
Sorry for posting so many of these! I'm finding them a little tricky and keep running into problems (Angry)

I've done two proofs for this, one uses $\epsilon$ and $\delta$, the other uses sequences.

The epsilon-delta version:

I need to find $|f(x)-f(x_0)|<\epsilon$ when $0<|x-x_0|< \delta$.
Since i'm trying to find continuity at 0, let $x_0=0$.

Therefore I need to find $|f(x)|< \epsilon$ when $0<|x|< \delta$.

For $x>0$, $|f(x)|< \epsilon \Leftrightarrow \ |sin(x)|< \epsilon \Leftrightarrow \ |x|< \epsilon$. Hence pick $\delta=\epsilon$.

For $x<0$, $|f(x)|<\epsilon \Leftrightarrow \ |x^2|< \epsilon \Leftrightarrow \ |x|< +\sqrt{\epsilon}$. Hence pick $\delta=+\sqrt{\epsilon}$

Therefore choose $\delta=min \{\epsilon, +\sqrt{\epsilon} \}$
$\color{red}{Is \ this \ the \ correct \ way \ to \ write \ \delta?}$

That was preliminary work so I have to write out the proper proof:

Let $\epsilon>0$ and define $\delta:=min \{ \epsilon, +\sqrt{\epsilon} \}$. When $0<|x|< \delta$:

$|f(x)-f(x_0)|=|x^2|< (\sqrt{\epsilon})^2=\epsilon$ for $x<0$.
$|f(x)-f(x_0)|=|sin(x)| \leq|x|< \epsilon$ for $x>0$.
Hence f is continuous $\forall x \in \mathbb{R}$.

Let $0 < \delta < 1$. If $|x| < \delta$ then $|\sin x| \leq |x|$ and $|x^2| \leq |x|$.

Therefore, for $\epsilon > 0$ let $\delta = \min \left( \epsilon, \tfrac{1}{2} \right)$.
• Mar 19th 2009, 10:22 AM
Showcase_22
ah, okay then.

Why is $
\delta = \min \left( \epsilon, \tfrac{1}{2} \right)
$
? Is it because you have $|x|< \delta<1$ so you have to choose a number between 0 and 1 exclusive?

For example, is $
\delta = \min \left( \epsilon, \ \frac{1}{3} \right)
$
also allowed?
• Mar 19th 2009, 08:35 PM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
ah, okay then.

Why is $
\delta = \min \left( \epsilon, \tfrac{1}{2} \right)
$
? Is it because you have $|x|< \delta<1$ so you have to choose a number between 0 and 1 exclusive?

For example, is $
\delta = \min \left( \epsilon, \ \frac{1}{3} \right)
$
also allowed?

That is fine. I just choose $\tfrac{1}{2}$ because I wanted $\delta < 1$.
• Mar 20th 2009, 02:31 AM
Showcase_22
ah great!

Can I also ask what you thought of my sequences proof? (Nerd)
• Mar 20th 2009, 09:17 AM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
Can I also ask hat you thught of my sequences proof? (Nerd)

I do not like it. To prove continuity by sequences you have to show that if $x_n \to 0$ then $f(x_n) \to 0$. The sequence does not need to stay positive or negative but it can just back and forth between positive and negative terms so you cannot just replace $f(x_n)$ by $x_n^2$ or $\sin (x_n)$ for all $n$.
• Mar 20th 2009, 10:57 AM
Showcase_22
umm, I don't suppose you could show me what you mean?

I thought the point was that there are two sequences that tend to the same limit- one from above and one from below. Then if you can show that when the function is applied to the limit of these sequences they produce the same value. This shows that a right hand limit exists, a left hand limit also exists and therefore the function is sequentially continuous.
• Mar 22nd 2009, 05:08 PM
xalk
Quote:

Originally Posted by Showcase_22
Sorry for posting so many of these! I'm finding them a little tricky and keep running into problems (Angry)

I've done two proofs for this, one uses $\epsilon$ and $\delta$, the other uses sequences.

The epsilon-delta version:

I need to find $|f(x)-f(x_0)|<\epsilon$ when $0<|x-x_0|< \delta$.
Since i'm trying to find continuity at 0, let $x_0=0$.

Therefore I need to find $|f(x)|< \epsilon$ when $0<|x|< \delta$.

For $x>0$, $|f(x)|< \epsilon \Leftrightarrow \ |sin(x)|< \epsilon \Leftrightarrow \ |x|< \epsilon$. Hence pick $\delta=\epsilon$.

For $x<0$, $|f(x)|<\epsilon \Leftrightarrow \ |x^2|< \epsilon \Leftrightarrow \ |x|< +\sqrt{\epsilon}$. Hence pick $\delta=+\sqrt{\epsilon}$

Therefore choose $\delta=min \{\epsilon, +\sqrt{\epsilon} \}$
$\color{red}{Is \ this \ the \ correct \ way \ to \ write \ \delta?}$

That was preliminary work so I have to write out the proper proof:

Let $\epsilon>0$ and define $\delta:=min \{ \epsilon, +\sqrt{\epsilon} \}$. When $0<|x|< \delta$:

$|f(x)-f(x_0)|=|x^2|< (\sqrt{\epsilon})^2=\epsilon$ for $x<0$.
$|f(x)-f(x_0)|=|sin(x)| \leq|x|< \epsilon$ for $x>0$.
Hence f is continuous $\forall x \in \mathbb{R}$.

Thanks in advance to anyone who posts! (Rofl)

Showcase.

The above proof is 100% correct,δ is indeed the minimum of { ε, $\sqrt{\epsilon}$}.

To write your proof in a better style:

let ε>0 , then choose δ= min{ ε, $\sqrt{\epsilon}$} which implies $\delta\leq\epsilon$.............................................1

$\delta\leq\sqrt{\epsilon}$.................................................. ............2

Now let xεDf ( Df = domain of f) and |x|<δ (and not 0<|x|<δ as you write in your proof,because you want x to take the zero value).

And now we have two cases:

Case 1) For, xεDf and x<0 and |x|<δ ====> (by using (2)) $|x|<\sqrt{\epsilon}\Longrightarrow |x|^2<\epsilon\Longrightarrow |f(x)|<\epsilon$

Case 2) For,xεDf and $x\geq 0$ and |x|<δ =====>( by using (1)) $|x|<\epsilon\Longrightarrow |sin(x)|<\epsilon\Longrightarrow |f(x)|<\epsilon$.

Since $|sin(x)|\leq |x|$ for all x,belonging to the real Nos

But $x<0\vee x\geq 0$,

Hence xεDf and |x|<δ =====> |f(x)|=|f(x)-0|=|f(x)-f(0)|<ε.

Thus f is continuous at x=0

How can you explain that phenomenon that happens very often in mathematics and particularly in analysis,where one writes a proof and then he/she cannot check his/her own proof and be sure of its correctness???
• Mar 23rd 2009, 03:39 AM
Showcase_22
Thanks! I had never come across a situation where $\delta$ was the minimum of two functions of $\epsilon$ and that made me a little confused =S

Do you also think my limit proof is a little iffy?
Here's my second attempt:

Quote:

Prove that $
f(x)=\begin{cases} x^2 \ \ \ \ \ \ \ \ \ x<0\\ sin(x) \ \ \ \ x \geq0 \end{cases}
$
is continuous at the origin.
Let $a_n$ be any sequence such that $a_n \rightarrow 0+$

Therefore $|a_n| \rightarrow 0+$ and $-|a_n| \rightarrow 0-$.

As before, I need to show that $\lim_{|a_n| \rightarrow 0+}f(|a_n|)=\lim_{-|a_n| \rightarrow 0-}f(-|a_n|)$ if $\lim_{x \rightarrow 0} f(x)$ exists.

Hence:

$f(\lim_{|a_n| \rightarrow \infty} f(|a_n|)= sin(0)=0$ if $x \geq 0$.

$f(\lim_{-|a_n| \rightarrow \infty}f(-|a_n|)=0^2=0$ if $x<0$.

The limits are the same so the function is continuous at the origin.

Is this method better?
• Mar 23rd 2009, 05:00 PM
xalk
Quote:

Originally Posted by Showcase_22
Thanks! I had never come across a situation where $\delta$ was the minimum of two functions of $\epsilon$ and that made me a little confused =S

In your preliminaries in the ε-δ proof you wrote and i quote:

"For x>0, $|f(x)|<\epsilon\leftrightarrow |sin(x)|<\epsilon\leftrightarrow |x|<\epsilon$.Hence pick δ=ε.

For x<0, $|f(x)|<\epsilon\leftrightarrow |x^2|<\epsilon\leftrightarrow |x|<+\sqrt{\epsilon}$.Hence pick $\delta = +\sqrt{\epsilon}$"

Here you put δ=ε and $\delta = +\sqrt{\epsilon}$,instead of putting $\delta_{1} = \epsilon$ and $\delta_{2} = +\sqrt{\epsilon}$.

And so when you wrote let δ= min{ $\epsilon,+\sqrt{\epsilon}$},it was $\delta$ = min{ $\delta_{1},\delta_{2}$}.

So you have come across a situation where δ is a function of ε .Only that the function of ε is hidding behind the $\delta_{1},\delta_{2}$.

In your ε-δ proof what you actualy did is that:

You used an example to justify the general theorem:

$\lim_{x\rightarrow a}{f(x)} = f(a)\Longleftrightarrow (\lim_{x\rightarrow a^+}{f(x)} = f(a) = \lim_{x\rightarrow a^-}{f(x)})$,instead of using that general theorem to prove your example.

Now in proving that general theorem ,in the converse proof:

$(\lim_{x\rightarrow a^+}{f(x)} = f(a)= \lim_{x\rightarrow a^-}{f(x)})\Longrightarrow \lim_{x\rightarrow a}{f(x)}$,we come across two deltas $\delta_{1},\delta_{2}$

For the sequence part i am still working on it
• Mar 24th 2009, 02:43 AM
Showcase_22
hmm, I see that I did the first limits problem incorrectly since I proved it for only one set of sequences, not for all sets.

What was wrong with my second attempt?(Crying) I'm can't see anything wrong with it since I proved the function as continuous at the origin for all sequences that tend to zero.
• Mar 24th 2009, 06:06 PM
xalk
Quote:

Originally Posted by Showcase_22
Thanks! I had never come across a situation where $\delta$ was the minimum of two functions of $\epsilon$ and that made me a little confused =S

Do you also think my limit proof is a little iffy?
Here's my second attempt:

Let $a_n$ be any sequence such that $a_n \rightarrow 0+$

Therefore $|a_n| \rightarrow 0+$ and $-|a_n| \rightarrow 0-$.

As before, I need to show that $\lim_{|a_n| \rightarrow 0+}f(|a_n|)=\lim_{-|a_n| \rightarrow 0-}f(-|a_n|)$ if $\lim_{x \rightarrow 0} f(x)$ exists.

Hence:

$f(\lim_{|a_n| \rightarrow \infty} f(|a_n|)= sin(0)=0$ if $x \geq 0$.

$f(\lim_{-|a_n| \rightarrow \infty}f(-|a_n|)=0^2=0$ if $x<0$.

The limits are the same so the function is continuous at the origin.

Is this method better?

Again your proof is correct except perhaps a few typo mistakes .

To write it again in a better style ,here it is:

Let { $a_{n}$} be a sequence in interval [ 0,+ $\infty$) ,for $x\geq 0$, and $\lim_{n\rightarrow\infty}{a_{n}}=0$.Then $\lim_{n\rightarrow\infty}{|a_{n}|}=0$ ,and { $|a_{n}|$} is in [ 0,+ $\infty$).

Where $a_{n}\neq 0$ for all n belonging to N.

But { $-|a_{n}|$} belongs to the (- $\infty,0$) interval,for x<0, and $\lim_{n\rightarrow\infty}{-|a_{n}|}=0$.

Now since the functions ,x^2, and sin(x) are continuous over the real Nos R and thus at x=0 we have that:

$\lim_{n\rightarrow\infty}{-|a_{n}|^2} = (\lim_{n\rightarrow\infty}{-|a_{n}|})^2 =0^2 =0=f(0)$,and $\lim_{n\rightarrow\infty}{sin(-|a_{n}|)} = sin(\lim_{n\rightarrow\infty}{-|a_{n}|}) = sin(0)=0=f(0)$.

Hence f is continuous at x= 0.

Of course there are a lot of theorems involved in the above proof and i think it would be a very good exercise if you could check thru the theorems the steps of your proof .I did that myself ,but you know the risk of making a mistake is very easy.

And of course it would be easier if you used two different general sequences { $a_{n}$} ,{ $b_{n}$} belonging to the intervals [0,+00) and (-00, 0) respectively instead of using the same sequence and running to the risk of making unforeseen mistakes.
• Mar 25th 2009, 06:27 AM
Showcase_22
Quote:

And of course it would be easier if you used two different general sequences { $a_{n}$} ,{ $b_{n}$} belonging to the intervals [0,+00) and (-00, 0) respectively instead of using the same sequence and running to the risk of making unforeseen mistakes.
I initially tried doing it just like this but when I took the modulus of both $a_n$ and $b_n$ they became defined on the same interval. This caused a huge problem because I didn't have a right and left hand limit, only a right hand one.

Is the method practically the same but instead of $-|a_n|$ we have $-|b_n|$?
• Mar 25th 2009, 07:04 PM
xalk
Quote:

Originally Posted by Showcase_22
I initially tried doing it just like this but when I took the modulus of both $a_n$ and $b_n$ they became defined on the same interval. This caused a huge problem because I didn't have a right and left hand limit, only a right hand one.

Is the method practically the same but instead of $-|a_n|$ we have $-|b_n|$?

You do not have to take the modulus of either sequence.

All you have to do is:

Let { $a_{n}$} be in the (- $\infty$,0) and $\lim_{n\rightarrow\infty}{a_{n}}=0$.

Also let { $b_{n}$} be in [0,+ $\infty$) and $\lim_{n\rightarrow\infty}{b_{n}}=0$.

Now again the functions ,x^2, sin(x) are continuous we have:

$\lim_{n\rightarrow\infty}{(a_{n})^2}= (\lim_{n\rightarrow\infty}{a_{n}})^2 = 0^2 = 0 =f(0)$ ,and

$\lim_{n\rightarrow\infty}{sin(b_{n})}= sin(\lim_{n\rightarrow\infty}{b_{n}}) = sin(0)= 0 = f(0)$

Hence f is continuous at x=0 .

Note ,since { $a_{n}$} is in (- $\infty$,0) then $\lim_{n\rightarrow\infty}{a_{n}}=0$ is a limit from the left of ,0.

For the same reason( { $b_{n}$} is in [0,+ $\infty$)) , $\lim_{n\rightarrow\infty}{b_{n}}=0$ is a limit from the right of ,0