if `|x| <1 ` then `(L t)_(n->oo)(1+x)(1+x^2)(1+x^4)........(1+x^(2n))=`

if |x|<1 then (Lt)_(n rarr oo)(1+x)(1+x^(2))(1+x^(4))......*(1+x^(2n))=

f(x)=lim_(n rarr oo)(1+x)(1+x^(2))(1+x^(4))..........(1+x^(2^(n-1))) then

Lim it_(n rarr oo){(1+x)(1+x^(2))(1+x^(4))......(1+x^(2^(n)))}= if |x|<1 has the value equal to

If |x| lt 1" thn "underset(n to oo)"Lt "(1+x)(1+x^(2))(1+x^(4))....(1+x^(2n))=

If |x|<1 , then lim_(n rarr oo)(1+x)(1+x^(2))(1+x^(4))...(1+x^(2^n))=

Let f(x)=x+(1-x)x^(3)+(1-x)(1-x^(2))x^(3)+.........+(1-x)(1-x^(2))......*(1-x^(n-1))x^(n);(n>=4) then :

Let f(x)=x+(1-x)x^3+(1-x)(1-x^2)x^3+.........+(1-x)(1-x^2)........(1-x^(n-1))x^n;(n >= 4) then :

Let f(x) = underset(n rarr oo)("Lim")( 2x^(2n) sin""(1)/(x) +x)/(1+x^(2n)) then find underset( x rarr -oo)("Lim") f(x)