8" sf "x_(1)=3 varepsilon_(2)x_(n+1)=sqrt(2+x_(n)),quad _(n rarr oo)^(1)x_(n)

If x_(1)=3 and x_(n+1)=sqrt(2+x_(n)),n<=1, then lim_(n rarr oo)x_(n) is

If x_(1)=2 and x_(n+1)=sqrt(x_(n)^(2)+8) , then x_(4)=

Let x_(0)=2(cos)(pi)/(6) and x_(n)=sqrt(2+x_(n-1))*n=1,2,3 find lim_(n rarr oo)2^(n+1)*sqrt(2-x_(n))

lim_(x rarr oo) (1+2/n)^(2n)=

If x_(1)=3 and x_(n+1)=sqrt(2+x_(n))" ",nge1, then lim_(ntooo) x_(n) is

If x_(1)=3 and x_(n+1)=sqrt(2+x_(n))" ",nge1, then lim_(ntooo) x_(n) is

If quad sqrt(3) and x_(n+1)=(x_(n))/(1+sqrt(1+x_(n)^(2))),AA n in N then lim_(n rarr oo)2^(n)x_(n) is equal to ,AA n in N

Let x_(1)=1 and x_(n+1)=(4+3x_(n))/(3+2x_(n)) for n>=1 If lim_(x rarr oo)x_(n) exists finitely,then the limit is equal to

let f(x)=lim_(n rarr oo)(x^(2n)-1)/(x^(2n)+1)

If x_(1)=3 and x_(n+1)=sqrt(2+x_(n)),n>=1, then backslash(lim backslash)_(x rarr oo)x_(n) is (a) -1 (b) 2(c)sqrt(5)(d)3