`("lim")_(xvec 0)((2^m+x)^(1/m)-(2^n+x)^(1/n))/xi se q u a lto` `1/(m2^m)-1/(n2^n)` (b) `1/(m2^m)+1/(n2^n)` `1/(m2^(-m))-1/(n2^(-n))` (d) `1/(m2^(-m))+1/(n2^(-n))`

lim_(xto0) ((2^(m)+x)^(1//m)-(2^(n)+x)^(1//n))/(x) is equal to

("lim")_(xvecoo){1/(1-n^2)+2/(1-n^2)+......n/(1-n^2)}i se q u a lto 0 (b) -1/2 (c) 1/2 (d) none of these

lim_(n to oo){(1^(m)+2^(m)+3^(m)+...+ n^(m))/(n^(m+1))} equals

If (2^(m+n))/(2^(m-n))=16 and a=2^(1/10) then ((a^(2m+n-p))^2)/((a^(m-2n+2p))^-1)=

2{(m-n)/(m+n)+1/3((m-n)/(m+n))^(3)+1/5((m-n)/(m+n))^(5) +..} is equals to

sum_(r=0)^n(-1)^r^n C_r[1/(2^r)+3/(2^(2r))+7/(2^(3r))+(15)/(2^(4r))+ u ptomt e r m s]= (2^(m n)-1)/(2^(m n)(2^n-1))

Show that : ((a^(m))/(a^(-n)))^(m-n)xx((a^(n))/(a^(-1)))^(n-1)xx((a^(l))/(a^(-m)))^(l-m)=1

IfI(m , n)=int_0^1x^(m-1)(1-x)^(n-1)dx ,(m , n in I ,m ,ngeq0),t h e n (a) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m-n))dx (b) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m+n))dx (c) I(m , n)=int_0^oo(x^(n-1))/((1+x)^(m+n))dx (d) I(m , n)=int_0^oo(x^n)/((1+x)^(m+n))dx

Evaluate : (2^(n)xx6^(m+1)xx10^(m-n)xx15^(m+n-2))/(4^(m)xx3^(2m+n)xx25^(m-1))

Given that lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2 , then evaluate : lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n) .