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#### DeusAbscondus

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- Jun 30, 2012

- 176

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

- Thread starter DeusAbscondus
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- Jun 30, 2012

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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

How areHello, 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

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- #3

- Jun 30, 2012

- 176

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

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- #4

- Jun 30, 2012

- 176

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...

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.

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

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.

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- #6

- Jun 30, 2012

- 176

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.

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

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- #7

- Jan 26, 2012

- 4,055

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.

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.

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- Jan 26, 2012

- 890

Long winded, I would just observe that: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.

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

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

CB

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- #9

- Jun 30, 2012

- 176

This might have nailed it for me Jameson, so thanks kindly.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.

But you agree, I take it, that it

Deus'Abs

My latest crumb of joy,

- Feb 13, 2012

- 1,704

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

a) the correct latin expression is

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, $\displaystyle \lim_{x \to 0}\frac{1}{x}$Two minor details...

a) the correct latin expression isin 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

$\chi$ $\sigma$

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- #12

- Jun 30, 2012

- 176

Further, $\displaystyle \lim_{x \to 0}\frac{1}{x}$does not existbecause 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 right, ProveIt?

Or are "the others" just wrong.

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

- Thread starter
- #13

- Jun 30, 2012

- 176

Thanks for the grammar and the calc inputTwo minor details...

a) the correct latin expression isin 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

$\chi$ $\sigma$

(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)

- Jan 29, 2012

- 1,151

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.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:isthis 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, itseems, 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)

As for the Khan Academy he says, in Limit Examples (part 2) | Calculus | Khan Academy, of exactly this problem that the limit

- Feb 13, 2012

- 1,704

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…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)

Kind regards

$\chi$ $\sigma$

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- #16

- 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