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Define limit of a function

Petrus

Well-known member
Feb 21, 2013
739
Define the limit of a function as \(\displaystyle x->+\infty\) and when \(\displaystyle x->a\)
(Only the case \(\displaystyle x->+\infty\) need to be treated.
I am suposed to do the delta epsilone proof for \(\displaystyle x->+\infty\) and for \(\displaystyle x->a\) but I think I missunderstand the question or do I?

Regards,
\(\displaystyle |\pi\rangle\)
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Define the limit of a function as \(\displaystyle x->+\infty\) and when \(\displaystyle x->a\)
(Only the case \(\displaystyle x->+\infty\) need to be treated.
I am suposed to do the delta epsilone proof for \(\displaystyle x->+\infty\) and for \(\displaystyle x->a\) but I think I missunderstand the question or do I?

Regards,
\(\displaystyle |\pi\rangle\)
I think they just want you to state the $\epsilon-\delta$ definitions for limits as $x\to\infty$ and $x\to a$; you can't really prove the statements since they're definitions after all. Unless they gave you a specific problem (i.e. show that $\lim\limits_{x\to \infty}x^2=\infty$), I don't see a need for you to prove anything.

I hope this helps!
 

Petrus

Well-known member
Feb 21, 2013
739
I think they just want you to state the $\epsilon-\delta$ definitions for limits as $x\to\infty$ and $x\to a$; you can't really prove the statements since they're definitions after all. Unless they gave you a specific problem (i.e. show that $\lim\limits_{x\to \infty}x^2=\infty$), I don't see a need for you to prove anything.

I hope this helps!
Thanks!
It Also say to proof sum,product and Quotient rule but end with (only the case \(\displaystyle x->+\infty\) need to be treated) What do they mean with that..?
for \(\displaystyle x->a\) My explain:
let f be real valued function that is defined in a open interval around a point a \(\displaystyle a \in \mathbb{R}\) without the point a. Then we say limits for f then x goes against x is equal to L and write as
\(\displaystyle \lim_{x->a}=L\)
for each \(\displaystyle \epsilon > 0 \) there exist a \(\displaystyle \delta > 0 \) such that
\(\displaystyle 0<|x-a|<\delta => |f(x)-L|< \epsilon\)
notice that \(\displaystyle |x-a|\) is the distance between the point x and the point a and \(\displaystyle |f(x)-L|\) is the distance between the number f(x) and L is this correct explain?

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Thanks!
It Also say to proof sum,product and Quotient rule but end with (only the case \(\displaystyle x->+\infty\) need to be treated) What do they mean with that..?
for \(\displaystyle x->a\) My explain:
let f be real valued function that is defined in a open interval around a point a \(\displaystyle a \in \mathbb{R}\) without the point a. Then we say limits for f then x goes against x is equal to L and write as
\(\displaystyle \lim_{x->a}=L\)
for each \(\displaystyle \epsilon > 0 \) there exist a \(\displaystyle \delta > 0 \) such that
\(\displaystyle 0<|x-a|<\delta => |f(x)-L|< \epsilon\)
notice that \(\displaystyle |x-a|\) is the distance between the point x and the point a and \(\displaystyle |f(x)-L|\) is the distance between the number f(x) and L is this correct explain?

Regards,
\(\displaystyle |\pi\rangle\)
For starters, that's how I would define $\displaystyle\lim_{x\to a}f(x)=L$. On the other hand, we have 3 cases to consider for infinite limits:

1. We have $\displaystyle \lim_{x\to a}f(x)=\pm\infty\iff \forall\,E>0,\,\exists\,\delta>0: 0<|x-a|<\delta\implies |f(x)|> E$.

2. We have $\displaystyle \lim_{x\to \pm\infty}f(x)=L \iff \forall\,\epsilon>0,\,\exists\, D>0: |x|>D \implies |f(x)-L|<\epsilon$

3. We have $\displaystyle\lim_{x\to \pm\infty}f(x)=\pm\infty \iff \forall\, E>0,\,\exists\, D>0: |x|>D \implies |f(x)|>E$

[HR][/HR]
With all that said, to me, it seems like you're supposed to use one of these definitions to prove:

a. The sum/difference rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)\pm g(x) = L\pm M$ (or the equivalent form for $x\to\infty$).

b. The product rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)g(x) = LM$ (or the equivalent form for $x\to \infty$).

c. The quotient rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M\neq 0$, then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M}$ (or the equivalent form for $x\to\infty$).


[HR][/HR]
I'll show product rule in the case of $x\to \infty$. So, we know that
\[\lim_{x\to\infty}f(x)=L\iff \forall\,\epsilon>0,\exists\,D_1>0: x>D_1\implies |f(x)-L|<\epsilon\]
and
\[\lim_{x\to\infty}g(x)=M\iff \forall\,\epsilon>0,\exists\,D_2>0: x>D_2\implies |g(x)-M|<\epsilon\]
(we will come back to this part in a bit because we're going to need to use each of these guys twice.)

Ideally, we would like to see the product $(f(x)-L)(g(x)-M)$ appear in the $|f(x)g(x)-LM|$ term. To get it to appear, we note that $(f(x)-L)(g(x)-M)=f(x)g(x)-Mf(x)-Lg(x) + LM$ and thus

\[\begin{aligned}f(x)g(x)-LM &= (f(x)-L)(g(x)-M) + Mf(x) + Lg(x) - 2LM\\ &= (f(x)-L)(g(x)-M) + Mf(x)-LM + Lg(x) - LM\\ &= (f(x)-L)(g(x)-M) + M(f(x)-L) + L(g(x)-M)\end{aligned}\]

Therefore,

\[|f(x)g(x)-LM| \leq \color{red}{|f(x)-L|}\color{blue}{|g(x)-M|} + |M|\color{green}{|f(x)-L|} + |L|\color{purple}{|g(x)-M|}.\]

Now, we use the facts that $\displaystyle\lim_{x\to\infty}f(x)=L$ and $\displaystyle\lim_{x\to \infty}g(x)=M$ to see that

\[\forall\,\epsilon>0,\,\exists \color{red}{D_1}>0: x>D_1 \implies \color{red}{|f(x)-L|}<\color{red}{\sqrt{\frac{\epsilon}{3}}}\]

\[\forall\,\epsilon>0,\,\exists \color{blue}{D_2}>0: x>D_2 \implies \color{blue}{|g(x)-M|}<\color{blue}{\sqrt{\frac{\epsilon}{3}}}\]

\[\forall\,\epsilon>0,\,\exists \color{green}{D_3}>0: x>D_3 \implies \color{green}{|f(x)-L|}<\color{green}{\frac{\epsilon}{3|M|}}\]

\[\forall\,\epsilon>0,\,\exists \color{purple}{D_4}>0: x>D_4 \implies \color{purple}{|g(x)-M|}<\color{purple}{\frac{\epsilon}{3|L|}}\]

Therefore, if we take $D=\max\{D_1,D_2,D_3,D_4\}$, we now see that

\[\forall\,\epsilon>0,\,\exists\,D>0:x>D\implies |f(x)g(x)-LM| < \color{red}{\sqrt{\frac{\epsilon}{3}}} \color{blue}{\sqrt{\frac{\epsilon}{3}}} + |M|\color{green}{\frac{\epsilon}{3|M|}} + |L|\color{purple}{\frac{\epsilon}{3|L|}} = \frac{\epsilon}{3}+ \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon.\]

Hence, $\displaystyle\lim_{x\to\infty} f(x)g(x) = \lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}g(x) = LM$.

I hope this makes sense! (Sun)
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks! Now I see what I did missunderstand.. I am well aware of those rule but not that they had same name.. I first thought they meant \(\displaystyle f(x)g(x)=f'(x)g(x)+f(x)g'(x)\) the derivate rule.. As My teacher Said they Dont expect well explain,
Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
For starters, that's how I would define $\displaystyle\lim_{x\to a}f(x)=L$. On the other hand, we have 3 cases to consider for infinite limits:

1. We have $\displaystyle \lim_{x\to a}f(x)=\pm\infty\iff \forall\,E>0,\,\exists\,\delta>0: 0<|x-a|<\delta\implies |f(x)|> E$.

2. We have $\displaystyle \lim_{x\to \pm\infty}f(x)=L \iff \forall\,\epsilon>0,\,\exists\, D>0: |x|>D \implies |f(x)-L|<\epsilon$

3. We have $\displaystyle\lim_{x\to \pm\infty}f(x)=\pm\infty \iff \forall\, E>0,\,\exists\, D>0: |x|>D \implies |f(x)|>E$

[HR][/HR]
With all that said, to me, it seems like you're supposed to use one of these definitions to prove:

a. The sum/difference rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)\pm g(x) = L\pm M$ (or the equivalent form for $x\to\infty$).

b. The product rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)g(x) = LM$ (or the equivalent form for $x\to \infty$).

c. The quotient rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M\neq 0$, then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M}$ (or the equivalent form for $x\to\infty$).


[HR][/HR]
I'll show product rule in the case of $x\to \infty$. So, we know that
\[\lim_{x\to\infty}f(x)=L\iff \forall\,\epsilon>0,\exists\,D_1>0: x>D_1\implies |f(x)-L|<\epsilon\]
and
\[\lim_{x\to\infty}g(x)=M\iff \forall\,\epsilon>0,\exists\,D_2>0: x>D_2\implies |g(x)-M|<\epsilon\]
(we will come back to this part in a bit because we're going to need to use each of these guys twice.)

Ideally, we would like to see the product $(f(x)-L)(g(x)-M)$ appear in the $|f(x)g(x)-LM|$ term. To get it to appear, we note that $(f(x)-L)(g(x)-M)=f(x)g(x)-Mf(x)-Lg(x) + LM$ and thus

\[\begin{aligned}f(x)g(x)-LM &= (f(x)-L)(g(x)-M) + Mf(x) + Lg(x) - 2LM\\ &= (f(x)-L)(g(x)-M) + Mf(x)-LM + Lg(x) - LM\\ &= (f(x)-L)(g(x)-M) + M(f(x)-L) + L(g(x)-M)\end{aligned}\]

Therefore,

\[|f(x)g(x)-LM| \leq \color{red}{|f(x)-L|}\color{blue}{|g(x)-M|} + |M|\color{green}{|f(x)-L|} + |L|\color{purple}{|g(x)-M|}.\]

Now, we use the facts that $\displaystyle\lim_{x\to\infty}f(x)=L$ and $\displaystyle\lim_{x\to \infty}g(x)=M$ to see that

\[\forall\,\epsilon>0,\,\exists \color{red}{D_1}>0: x>D_1 \implies \color{red}{|f(x)-L|}<\color{red}{\sqrt{\frac{\epsilon}{3}}}\]

\[\forall\,\epsilon>0,\,\exists \color{blue}{D_2}>0: x>D_2 \implies \color{blue}{|g(x)-M|}<\color{blue}{\sqrt{\frac{\epsilon}{3}}}\]

\[\forall\,\epsilon>0,\,\exists \color{green}{D_3}>0: x>D_3 \implies \color{green}{|f(x)-L|}<\color{green}{\frac{\epsilon}{3|M|}}\]

\[\forall\,\epsilon>0,\,\exists \color{purple}{D_4}>0: x>D_4 \implies \color{purple}{|g(x)-M|}<\color{purple}{\frac{\epsilon}{3|L|}}\]

Therefore, if we take $D=\max\{D_1,D_2,D_3,D_4\}$, we now see that

\[\forall\,\epsilon>0,\,\exists\,D>0:x>D\implies |f(x)g(x)-LM| < \color{red}{\sqrt{\frac{\epsilon}{3}}} \color{blue}{\sqrt{\frac{\epsilon}{3}}} + |M|\color{green}{\frac{\epsilon}{3|M|}} + |L|\color{purple}{\frac{\epsilon}{3|L|}} = \frac{\epsilon}{3}+ \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon.\]

Hence, $\displaystyle\lim_{x\to\infty} f(x)g(x) = \lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}g(x) = LM$.

I hope this makes sense! (Sun)
Hi,
I got some question, what means with \(\displaystyle D_1\) is it like delta? then I don't see how you got \(\displaystyle \sqrt{\frac{\epsilon}{3}}\) and \(\displaystyle \frac{\epsilon}{3|M|}\), \(\displaystyle \frac{\epsilon}{3|L|}\), what does this exactly mean \(\displaystyle D=\max\{D_1,D_2,D_3,D_4\}\). Your explaining was really good and it did make sense just some question and then evrything is clear! Thanks for taking your time!

Regards,
\(\displaystyle |\pi\rangle\)
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Hi,
I got some question, what means with \(\displaystyle D_1\) is it like delta? then I don't see how you got \(\displaystyle \sqrt{\frac{\epsilon}{3}}\) and \(\displaystyle \frac{\epsilon}{3|M|}\), \(\displaystyle \frac{\epsilon}{3|L|}\), what does this exactly mean \(\displaystyle D=\max\{D_1,D_2,D_3,D_4\}\). Your explaining was really good and it did make sense just some question and then evrything is clear! Thanks for taking your time!

Regards,
\(\displaystyle |\pi\rangle\)
With regards to the $D$'s, I suppose you can think of them as $\delta$'s, but in a different way. When you're doing $x\to a$ limits, you take $\delta$ to be a value relatively close to $a$, but for $x\to\infty$ limits, $D$ is usually assumed to be some very large number on the real line where we seem to see convergence (i.e. $|f(x)-L|<\epsilon$) for $x>D$.

With regards to my choices for $\epsilon$'s: when we had the inequality

\[|f(x)g(x)-LM|\leq |f(x)-L||g(x)-M| + |M||f(x)-L| + |L||g(x)-M|\]

note that we have three different terms being added together. If we want $|f(x)g(x)-LM|<\epsilon$, we would like each piece to be less than $\dfrac{\epsilon}{3}$. With that, we must have

\[|f(x)-L||g(x)-M| < \frac{\epsilon}{3},\quad |M||f(x)-L| < \frac{\epsilon}{3},\quad\text{and}\quad |L||g(x)-M|<\frac{\epsilon}{3}\]

Thus, for $\displaystyle|f(x)-L||g(x)-M| < \frac{\epsilon}{3}$, we would need $\displaystyle |f(x)-L| <\sqrt{\frac{\epsilon}{3}}$ and $\displaystyle |g(x)-M| < \sqrt{\frac{\epsilon}{3}}$.

For $\displaystyle |M||f(x)-L| < \frac{\epsilon}{3}$, we would need $\displaystyle |f(x)-L| < \frac{\epsilon}{3|M|}$.

For $\displaystyle |L||g(x)-M| < \frac{\epsilon}{3}$, we would need $\displaystyle |g(x)-M| < \frac{\epsilon}{3|L|}$.

The crucial thing to note here is that we have all these inequalities satisfied for different $D$ values!

Since there were 4 total pieces in that sum (but three terms that were being added together), we had to have four different $D$ values: $D_1$, $D_2$, $D_3$ and $D_4$. Thus, if we have to have $x>D_1$ and $x>D_2$ and $x>D_3$ and $x>D_4$ in order to have

\[x>D\implies |f(x)g(x)-LM|< \sqrt{\frac{\epsilon}{3}}\sqrt{\frac{\epsilon}{3}} + |M|\frac{\epsilon}{3|M|} + |L|\frac{\epsilon}{3|L|} = \frac{\epsilon}{3}+\frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\]

we take $D$ to be the largest of the $D_i$'s; i.e. $D=\max\{D_1, D_2, D_3, D_4\}$ (Note here that if we were dealing with multiple $\delta$'s [say $\delta_1,\ldots, \delta_n$], we would then take $\delta$ to be the smallest of the $\delta_i$'s; i.e. $\delta=\min \{\delta_1,\ldots,\delta_n\}$).

With all that said, I hope this clarifies things!! (Sun)
 

Petrus

Well-known member
Feb 21, 2013
739
Hello,
i try find the proof for all of the rest of those law when \(\displaystyle x->\infty\) but can't find them especially Quotient rule as I think the rest I got:S idk if it's correct. Anyone got a Link for them all proof?

Regards,
\(\displaystyle |\pi\rangle\)
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Hello,
i try find the proof for all of the rest of those law when \(\displaystyle x->\infty\) but can't find them especially Quotient rule as I think the rest I got:S idk if it's correct. Anyone got a Link for them all proof?

Regards,
\(\displaystyle |\pi\rangle\)
I was going to respond to this a couple days ago, but then got distracted by something else and forgot about it... (Wasntme)

You can find all of the other proofs here.