Evaluate: `("lim")_(nvecoo)((a1 1/x+a2 1/x++a n1/x)/n)^(n x)`

Evaluate: ("lim")_(nvecoo)[1/(n^2)sec^2 1/(n^2)+2//n^2sec^2 4/(n^2)++1/nsec^2 1]

Evaluate: ("lim")_(nvecoo)n^(-n^2){(n+2^0)(n+2^(-1))(n+2^(-2))(n+2^(-n+1))}^n

Evaluate: ("lim")_(nvecoo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))

Evaluate ("lim")_(nvecoo){cos(x/2)cos(x/4)cos(x/8) cos(x/(2^n))}

Evaluate: ("lim")_(nvecoo)(sumr=1nsqrt(r)sumr=1n1/(sqrt(r)))/(sumr=1n r)

Show that ("lim")_(nvecoo)(1/(n+1)+1/(n+2)++1/(6n))=log6

Evaluate: ("lim")_(nvecoo)((1^2+2^2+3^3++n^2)(1^3+2^3+3^3++n^3))/(1^6+2^6+3^6++n^6)

If [x] denotes the greatest integer less than or equal to x , then evaluate ("lim")_(ntooo)1/(n^2)([1.x]+[2.x]+[3.x]+....+[n.x])

Find the values of a if f(x)=("lim")_(n rarr oo)(a x^(2n)+2)/(x^(2n)+a+1) is continuous at x=1.

If f(x) is integrable over [1,], then int_1^2f(x)dx is equal to ("lim")_(nvecoo)1/nsum_(r=1)^nf(r/n) ("lim")_(nvecoo)1/nsum_(r=n+1)^(2n)f(r/n) ("lim")_(nvecoo)1/nsum_(r=1)^nf((r+n)/n) ("lim")_(nvecoo)1/nsum_(r=1)^(2n)f(r/n)