JEE Maths Zone - 2019-08-27 (page 1 of 2)

Even looking in extended real number. You can define a symbol infinity, holding the following properties. wikimedia.org/api/rest_v1/media/math/render/svg/…

Jasmine

@Mann there is a difference between does not exist and indeterminate right

@Mann yes

Mann

Umm, let me verify that. I think there shouldn't be.

I am not really sure about the difference of this two term. But I am assuming it wouldn't matter. In any case, If I do think It will I will try to address

Jasmine

Please tell me mathjax code for limit

Mann

$1^{\infty}$ is not defined even in extended real numbers.

This is what I am saying.

Jasmine

@Mann yes

Mann

3:36 PM

You can't derive it using the properties in the image. And all the properties about real number you already know

Jasmine

53 secs ago, by Jasmine

Please tell me mathjax code for limit

Mann

Okay, now say the next question.

Advil Sell

@Jasmine \lim_{n \to X} $\lim_{n \to X}$

Mann

\lim_{x \to \infty} x^{\frac{1}{x}-1}

Jasmine

@AdvilSell Thank you !

What is $\lim_{x \to 1-} x^{\infty↑}$

Mann

3:39 PM

@Jasmine What's the next query following that?

$x^{\infty}$ is again not defined for any $x$

Advil Sell

@Mann isn't (1/2)^{Infinity} = 0 ?

Mann

Does the limit for x, applied after the limit to infinity?

Jasmine

@Mann please see edited question

Mann

There are two possible meaning of that expression.

1) Order : lim x lim n x^n in which case it is 0.

2) Order: lim n lim x in which case it is 1

@AdvilSell no

In general the order of limits applied matter

My presumption was that the limits of n and m were applied first. Then x was applied.

Advil Sell

@Mann okay so $$\lim_{n \to infi} (1/2)^n=0$$ ?

Mann

3:43 PM

Yep @AdvilSell

@Jasmine is it okay now?

Advil Sell

@Mann oh do it follow from the fact that Infinity is not a number ? But a symbol ?

Mann

@AdvilSell yes. But infact, everything is a symbol only.

Symbols satisfying certain properties.

It's not like the numbers have a special status.

They are also symbols satisfying certain properties

Jasmine

@Mann ↓

6 mins ago, by Jasmine

What is $\lim_{x \to 1-} x^{\infty↑}$

Mann

I said na.

There are two possible meaning of that expression.
1) Order : lim x lim n x^n in which case it is 0.
2) Order: lim n lim x in which case it is 1

There is no notion of both the limits being applied at same time.

You have to define the order of limits

Advil Sell

Hmm...what makes Infinity different ? I mean how to derive the above questions (1/2^{infinity} one ) using properties of infinity ?

Mann

3:47 PM

You can't maybe derive it. We don't know that, I don't, so far.

Jasmine

@Mann and what is $\lim_{x \to 1+ } x ^{\infty ↑} $

Mann

Nothing makes infinity different, it's just a new object satisfying new rules added to the already existing set of symbols inside the real number.

Two meanings again
1) Lim x lim n in which case "towards infinity"
2) lim n lim x in which case 1.

Advil Sell

@Mann then why is (1/2)^1 = 1/2 and (1/2)^{infinity} is undefined ?

Jasmine

So if we calculate $\lim_{x\to 1} x^{\infty↑}$ we say LHL not equal RHL

That means Limit doesn't exist

Mann

Because we have properly defined what all those notations means in case of a^b where a and b are real numbers.

"So if we calculate limx→1x∞↑ we say LHL not equal RHL"
This depends upon the interpretation

Jasmine

3:51 PM

Now if does not exist and indeterminate was are not equal then

Mann

But if you agree to use lim x lim n f(n,x). Then yes

Jasmine

@Mann I dont get it

Mann

Which part?

Advil Sell

@Mann so are you saying we haven't defined infinity properly ?

Jasmine

I am applying limit on x first

Mann

3:53 PM

Why?

Jasmine

@Mann then yes means

Mann

@AdvilSell that discussion will tend to a highly advanced topic but it is more or less defined enough. It's just the notation $a^{\infty}$ is not defined.

@Jasmine I can explain why did I use my interpretation but it will take some time

Following the order lim n lim x f(n,x) is not the best option here

Jasmine

@Mann you said yes on what

@Mann consider limit on x first

Mann

Then yea, the limit exists and is equal.

limx→1x∞↑

For this

Jasmine

In the case 1↑ to the power infinity ↑

Advil Sell