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

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4:00 PM

Then what you are asking is this I assume :

$\lim_{n \to \infty } \lim_{x \to 1} x^n $

Right?

@Mann yes

Then that's just equal to $\lim_{n \to \infty} 1^n $

Because for any finite $n$, the limit for $\lim_{x \to 1} x^n$ exists

@Mann that doesn't exist or is indeterminate

What doesn't exist or is indeterminate?

@Mann what is its value

4:03 PM

Whose value?

I think its indeterminate only when we are working with same variables

$\lim_{n \to \infty} 1^n$

?

@Mann yes

What is its value

For any finite $n$, $1^n = 1 $

By using the definition of limits, we can say that the limit exists and is 1

@Mann 1 is not exact 1

4:04 PM

What, do you mean?

I mentioned x tends to 1

Yeaa but

I already took the limit

$\lim_{n \to \infty } (\lim_{x \to 1} x^n)$

I took the limit of $x$ tending to one already

And it gave us the exact limiting value as output

@Mann so you mean 'n tending to infinity' is 'finite n'

That limiting value is indeed $1^n$

No, It isn't

Okayy wai

t

This is what I am doing

$\lim_{n \to \infty } (\lim_{x \to 1} x^n) = 1$

@Mann so this is 1 you mean

4:07 PM

Yes

$\lim_{n \to \infty } (\lim_{x \to 1} x^n) = 1$ but $\lim_{x \to 1}(\lim_{n \to \infty } x^n) $ doesn't exist

@Mann is the 2nd term is not defined because we LHL not equal to RHL

I didn't get you?

By LHl I mean x tends to 1+ following the same order of limits

Ahhhhhh

DOn't look at it like that

Seee

@Mann why :(

4:11 PM

The meaning of the expression $\lim_{n\to \infty} x^n$ means that

Let $f_n(x)= x^n$

Then the limit function is defined $f(x) = \lim_{n \to \infty} f_n(x)$

How we evaluate this $f(x)$ is like this

We take a fixed $x_0$

for this $x_0$ we find $\lim_{n\to \infty} f_n(x_0)$

This we call $f(x_0)$

We repeated this procedure for all $x_0$ and come up with the function $f(x)$

Then on this newly found function $f(x)$

We take the $\lim_{x \to 1} f(x)$

Last question What is $ \lim_{x \to 1+} $ $\lim_{n \to {\infty} }x^n$

$\lim_{x \to 1_+} (\lim_{n \to {\infty}})x^n$

Okayy so first of all, this begs the questions

What is

$\lim_{n \to \infty} x^n$

Right?

I state that it is

1 min ago, by Jasmine

Last question What is $ \lim_{x \to 1+} $ $\lim_{n \to {\infty} }x^n$

What is this ^

Yeaa I wrote the same thing

"limx→1+(limn→∞)xn"

This is your question isn't it?

Ohh no

4:17 PM

Why not?

I am using wrong order of limits

First limit x and then limit n

?

In what

@Mann nvm I got everything

No I used first limit of n and m in the question you had given me. Then I used limit of x

cos and sin one

I was just messing up single variable limits with double variable limits

4:19 PM

Yeaa, both are different

You seem to be thinking about iterated limits

In multivariable calculus, an iterated limit is an expression of the form lim y → q ( lim x → p f ( x , y ) ) . {\displaystyle \lim _{y\to q}{\big (}\lim _{x\to p}f(x,y){\big )}.\,} One has an expression whose value depends on...

@AdvilSell

Now say

@Mann Ok I have got it Thank you :-)

Cool!

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Aug '1927

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