sequence

[swift crest]

{xn} is a sequence and limit (xn+1)/xn= lembda(<1) so what will be limit of xn?

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[swift crest]

1+1/xn

My answer is 1

[mighty flame]

I assume you've shown the limit of x_n exists already since you're not asking about that, so if the limit of x_n is L, then (L+1)/L=lambda, to which you get a quadratic in L

[swift crest]

Quadratic how?

You are taking both sides lambd?

[mighty flame]

sorry linear

$L+1=\lambda L$

[foggy vortex]

Omegabet_

[mighty flame]

had L+1/L in my mind instead of (L+1)/L

[olive blade]

is n+1 an index?

[swift crest]

L= 1/(1-lambda)

42

In hindi they gave nothing much information

[mighty flame]

1/(lambda-1)

$1=L(\lambda-1)$

[foggy vortex]

Omegabet_

[swift crest]

Yes

And lambda is less than 1

We will get minus quantity

[mighty flame]

L is negative, yes

[swift crest]

So what should I do with options?

[mighty flame]

pick the correct one...

[swift crest]

D

None