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[SOLVED] Simple limit question in attached pdf

DeusAbscondus

Active member
Jun 30, 2012
176
Hello, I have a simple question regarding limits.

I have attached this question as a pdf for reasons of accuracy (i can latex to pdf but haven't learnt how to work with Mathjax directly here)

Regs,
DeusAbsView attachment 289
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Hello, I have a simple question regarding limits.
I have attached this question as a pdf for reasons of accuracy (i can latex to pdf but haven't learnt how to work with Mathjax directly here)View attachment 289
How are "no limit" and "limit not defined" different?
 

DeusAbscondus

Active member
Jun 30, 2012
176
I'm just a beginner to all this, taking my first calculus course this semester.
I honestly didn't realize the terms were synonymous. I thought perhaps "undefined" had a technical meaning like: "no conditions under which this could be true" whereas, "No Limit" seems to connote "limitless" and, while working on a problem with $\infty$ all the way through it, well, I got confused.

Are you telling me that my working out is correct by the way?

thx for the interaction Plato
DeusAbs

PS I have been coming here for weeks and only just worked out how to put latex in a message box!
I got to show the world!$$\lim_{x\rightarrow0}\frac{1}{x}=\infty$$

Now that is trippy! :)
 

DeusAbscondus

Active member
Jun 30, 2012
176
Now that I have worked out how to use latex here, i may as well repeat my working out, Plato, and anyone else reading this, and ask for your comments as to its fullness and correctness or otherwise.

Thx kindly,
Deus'Abs

Here is the question in my text (the text is a collection of notes built up heterogeneously, over the years, with several contributors, and no editorial oversight as far as I can judge, and contains quite a few errors, unfortunately ... indeed, most annoyingly for a beginner such as myself)

$$\lim_{x\rightarrow \infty}\frac{x^{2}+6}{x-4}=\lim_{x\rightarrow \infty}\frac{(x^2/x^2)+(6/x^2)}{(x/x^2)-(4/x^2)}\\$$ (by substituting $\infty$ for any occurrence of x)
$$=\lim_{x\rightarrow \infty}\frac{1+0}{(0-0}$$
which is undefined.
But the answer given in the text is: ``No limit"
Can this be correct? Waiting with baited breath.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Now that I have worked out how to use latex here, i may as well repeat my working out, Plato, and anyone else reading this, and ask for your comments as to its fullness and correctness or otherwise.

Thx kindly,
Deus'Abs

Here is the question in my text (the text is a collection of notes built up heterogeneously, over the years, with several contributors, and no editorial oversight as far as I can judge, and contains quite a few errors, unfortunately ... indeed, most annoyingly for a beginner such as myself)

$$\lim_{x\rightarrow \infty}\frac{x^{2}+6}{x-4}=\lim_{x\rightarrow \infty}\frac{(x^2/x^2)+(6/x^2)}{(x/x^2)-(4/x^2)}\\$$ (by substituting $\infty$ for any occurrence of x)
$$=\lim_{x\rightarrow \infty}\frac{1+0}{(0-0}$$
which is undefined.
But the answer given in the text is: ``No limit"
Can this be correct? Waiting with baited breath.
I'd do it like this...

\[ \displaystyle \begin{align*} \lim_{x \to \infty}\frac{x^2 + 6}{x - 4} &= \lim_{x \to \infty}\frac{x^2 - 4x + 4x + 6}{x - 4} \\ &= \lim_{x \to \infty}\left[\frac{x(x - 4)}{x - 4} + \frac{4x + 6}{x - 4} \right] \\ &= \lim_{x \to \infty}\left( x + \frac{4x + 6}{x - 4}\right) \\ &= \lim_{x \to \infty}\left( x + \frac{4x - 16 + 22}{x - 4} \right) \\ &= \lim_{x \to \infty}\left[ x + \frac{4(x - 4)}{x - 4} + \frac{22}{x - 4}\right] \\ &= \lim_{x \to \infty}\left(x + 4 + \frac{22}{x - 4}\right) \\ &\to \infty \end{align*} \]

The reason I did it like this is because it's easier to see that the x term greatly overpowers the fractional term and therefore won't approach different values from either side.

It's possible to say that since this does not approach a finite value, you can say that "the limit does not exist". But it is also correct to say that the function will increase without bound, or tend to infinity.
 

DeusAbscondus

Active member
Jun 30, 2012
176
further explanation sort; graph provided

So, ProveIt, you agree with my original query: the answer given to this was "No Limit" and by that the authors did not mean: "Undefined" but that there is no limit to this function?

I have graphed it and it is a hyperbola and therefore, by definition, asymptotic in both directions. Please find attached a screenshot png of the graph with its first derivative.

Now I am completely confused!
Please explain explicitly what the authors of my text had in mind in providing No Limit as an answer, and, the difference between "No Limit" on the one hand and "Undefined" on the other.

Thanks,
deusAbs
PS: it is 1am where I live and my eyes are watering from tiredness; please respond and I will get your considered reply, with many thanks, tomorrow morning my time, when I can come at this with a freshened brain :)
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
You're absolutely correct that as \(\displaystyle x \rightarrow \infty, y \rightarrow \infty\). You've shown this to be be true through algebraic manipulation and confirmed this by graphing the function. At this point it comes down to semantics and normally in starting calculus courses students are told to write \(\displaystyle +\infty\) or \(\displaystyle -\infty\) when the function tends to one of these ends instead of just writing "undefined", because knowing the end behavior of a function can be very useful.

On the other hand, \(\displaystyle \lim_{x \rightarrow 4}f(x)\) is undefined for this function because on one side it is approaching positive infinity and on the other it is approaching negative infinity. If just looking at approaching from one side, or a one-sided limit, then these can be found.

When you have asymptotes diverging in opposite directions then you confidently answer that the limit does not exist and not need to clarify that statement anymore. If you write that a limit is undefined or d.n.e. which tends to infinity or negative infinity, then you could get into trouble for not stating this information as often the entire point of certain limits is this bit of information.
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
I'd do it like this...

\[ \displaystyle \begin{align*} \lim_{x \to \infty}\frac{x^2 + 6}{x - 4} &= \lim_{x \to \infty}\frac{x^2 - 4x + 4x + 6}{x - 4} \\ &= \lim_{x \to \infty}\left[\frac{x(x - 4)}{x - 4} + \frac{4x + 6}{x - 4} \right] \\ &= \lim_{x \to \infty}\left( x + \frac{4x + 6}{x - 4}\right) \\ &= \lim_{x \to \infty}\left( x + \frac{4x - 16 + 22}{x - 4} \right) \\ &= \lim_{x \to \infty}\left[ x + \frac{4(x - 4)}{x - 4} + \frac{22}{x - 4}\right] \\ &= \lim_{x \to \infty}\left(x + 4 + \frac{22}{x - 4}\right) \\ &\to \infty \end{align*} \]

The reason I did it like this is because it's easier to see that the x term greatly overpowers the fractional term and therefore won't approach different values from either side.

It's possible to say that since this does not approach a finite value, you can say that "the limit does not exist". But it is also correct to say that the function will increase without bound, or tend to infinity.
Long winded, I would just observe that:

\[\lim_{x \to \infty}\frac{x^2 + 6}{x - 4} = \lim_{x \to \infty}\frac{x + 6x^{-1}}{1 - 4x^{-1}} \sim x\]

and so the limit does not exist as \(\frac{x^2 + 6}{x - 4}\to \infty \) as \(x \to \infty \)

CB
 

DeusAbscondus

Active member
Jun 30, 2012
176
On the other hand, \(\displaystyle \lim_{x \rightarrow 4}f(x)\) is undefined for this function because on one side it is approaching positive infinity and on the other it is approaching negative infinity. If just looking at approaching from one side, or a one-sided limit, then these can be found.
This might have nailed it for me Jameson, so thanks kindly.
But you agree, I take it, that it is confusing to equate the two expressions: "No Limit" and "undefined", right?

Deus'Abs

My latest crumb of joy, in hac lacrimarum vale: $$\lim_{x\to0}\frac{1}{x}=\infty$$
 

chisigma

Well-known member
Feb 13, 2012
1,704
My latest crumb of joy, in hac lacrimarum vale: $$\lim_{x\to0}\frac{1}{x}=\infty$$
Two minor details...

a) the correct latin expression is in hac lacrimarum valle...

b) the correct math expression is $\displaystyle \lim_{x \rightarrow 0+} \frac{1}{x}= + \infty$ ,
$\displaystyle \lim_{x \rightarrow 0-} \frac{1}{x}= - \infty$

Kind regards

$\chi$ $\sigma$

 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Two minor details...

a) the correct latin expression is in hac lacrimarum valle...

b) the correct math expression is $\displaystyle \lim_{x \rightarrow 0+} \frac{1}{x}= + \infty$ ,
$\displaystyle \lim_{x \rightarrow 0-} \frac{1}{x}= - \infty$

Kind regards

$\chi$ $\sigma$

Further, $\displaystyle \lim_{x \to 0}\frac{1}{x}$ does not exist because the function approaches different values (or directions) from each side.
 

DeusAbscondus

Active member
Jun 30, 2012
176
Further, $\displaystyle \lim_{x \to 0}\frac{1}{x}$ does not exist because the function approaches different values (or directions) from each side.


Thanks, ProveIt.
Wow! have I ambled into an old mathematical controversy after just a few weeks studying calculus?
That's actually exciting.

No, really, in a spirit of scientific curiosity: is this a genuine area of controversy? I mean, Sal Khan, maths teacher at Khanacademy.com, endorses this limit as being "mind-blowingly transcendent .... or something" on one of his videos, and I don't think he is a mathematical dunce. So, it seems, on the face of it, to be controversial.

Is this right, ProveIt?
Or are "the others" just wrong.

( Btw, Hello from fellow aussie at Lismore, NSW)
 

DeusAbscondus

Active member
Jun 30, 2012
176
Two minor details...

a) the correct latin expression is in hac lacrimarum valle...

b) the correct math expression is $\displaystyle \lim_{x \rightarrow 0+} \frac{1}{x}= + \infty$ ,
$\displaystyle \lim_{x \rightarrow 0-} \frac{1}{x}= - \infty$

Kind regards

$\chi$ $\sigma$

Thanks for the grammar and the calc input
(i used to chant those words everyday around sunset but rarely if ever saw the words written; and never did much systematic Latin study; would if i had 2 or 3 lives to live though)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Thanks, ProveIt.
Wow! have I ambled into an old mathematical controversy after just a few weeks studying calculus?
That's actually exciting.

No, really, in a spirit of scientific curiosity: is this a genuine area of controversy? I mean, Sal Khan, maths teacher at Khanacademy.com, endorses this limit as being "mind-blowingly transcendent .... or something" on one of his videos, and I don't think he is a mathematical dunce. So, it seems, on the face of it, to be controversial.

Is this right, ProveIt?
Or are "the others" just wrong.

( Btw, Hello from fellow aussie at Lismore, NSW)
No, there is no controversy. Sometimes, if the limit from both sides is [tex]\infty[/tex], some people will say the limit is "infinity" others will say "does not exist". But here, because the limit from one side is [tex]\infty[/tex] and from the other [tex]-\infty[/tex] there is no question- the limit does not exist.

As for the Khan Academy he says, in Limit Examples (part 2) | Calculus | Khan Academy, of exactly this problem that the limit does not exist. Nothing controversial about that!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Thanks for the grammar and the calc input
(i used to chant those words everyday around sunset but rarely if ever saw the words written; and never did much systematic Latin study; would if i had 2 or 3 lives to live though)
The song ‘Salve Regina’ is probably the oldest holy choral music composition to this day executed. It was composed in XI century but who was its writer is controversial matter. The ‘tradition’ awards it to Hermann von Reichenau but Pope Gregorio VII, Saint Anselmo of Baggio, Saint Peter of Bezonzio and Saint Bernard of Charavalle are also quoted. Here the song, the latin and english text…

Salve regina misericordiae: vita, dulcedo, et spes nostra, salve.
Ad te clamamus, exules, filii Hevae.
Ad te suspiramus, gementes et flentes in hac lacrimarum valle.
Eia ergo, advocata nostra, illos tuos misericordes oculos ad nos converte.
Et Jesum, benedictum fructum ventris tui, nobis post hoc exilium ostende.
O clemens: O pia: O dulcis Virgo Maria.
Hail, O queen of mercy: our life, our sweetness and our hope, hail.
To you we cry, banished children of Eve.
To you we send up our sighs, mourning and weeping in this vale of tears.
Therefore turn, you our advocate, your merciful eyes towards us.
And after this our exile, show to us the blessed fruit of thy womb, Jesus.
O clement: O loving: O sweet Virgin Mary.

Salve_Regina.png
Today however, after the Vatican Council II, ‘Salve Regina’ is rarely adopted in catholic holy ceremonies and maintains purely ‘historical’ significance…

Kind regards

$\chi$ $\sigma$
 

DeusAbscondus

Active member
Jun 30, 2012
176
The song ‘Salve Regina’

Like my hero, Sam Beckett, I only use the biblical and religious language for its sublimine quality of expression.
Again like Beckett, I adore Dante, but not for the religious message -which I regard as conveying a defunct, morally wicked, humanly craven world-view- but for the sustained, beauteous, poetic vision.

Thanks for the sheet music though (i spent some time in my misguided youth as a monk with the Cistercians, and did, in fact, read these words and see this music every night, but it was 30 years ago and a whole life away, and more than one "world-shift" away too)

DeusAbscondus