Help..!! Limits..!!

Limitm→∞ Limitn→∞ (1+cos2m(n!πx)) where n! denotes factorial of n..

Answer=2 if x is rational and 1 if x is irrational..

4 Answers

1
chintan patel patel ·

-1 ≤ cos x ≤ 1
0 ≤ cos 2mk ≤ 1
for cos k =1
cos 2m k=1
we have soln = 2
and for other we have cos 2m k →0 therefore limiting value is 1
k=n!x
k should be of the form ,a*pie ;a belongs to N
have you understood
i am not able to explain properly
n not sure of this way

11
Khilen Khara ·

0 ≤ cos 2mk ≤ 1

How can u say dis??????

1
gordo ·

hint:
if x is rational, it can be written as (1+cos2mn!*pi*(p/q)) where p,q belong to N if not rational, it cannot be expressed as p/q
for each case see what type of a term you get inside the cos, and accordingly decide.
cheers!

1
aditya ravichandran ·

if x is rational

x=pq

and as n→∞

n > q

So, n!q=Integer

cos2(integer*Ï€) =1

hence limit will be 2

In the second case x is irrational

cos2(irrational)<1
lim m→∞cos2m(irrational)=0

hencce limit is 1

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