`lim_(x->oo) (x^m.e^-x)`

Solve lim_(x->oo)(e^x-e^-x)/(e^x-e^-x)

lim_(x->oo) ((x^2-2x+1)/(x^2-4x+2))^x is equal to

lim_(x->oo) (sinx/x) =

lim_(x -> oo) x^n / e^x = 0 , (n is an integer) for

lim_(x->oo)(e^(11x)-7x)^(1/(3x))

lim_(x->oo)(x^2 - sinx)/(x^2-2)

lim_(x->oo)(1-x+x.e^(1/n))^n

If f(x) = lim_(n->oo) tan^(-1) (4n^2(1-cos(x/n))) and g(x) = lim_(n->oo) n^2/2 ln cos(2x/n) then lim_(x->0) (e^(-2g(x)) -e^(f(x)))/(x^6) equals

If lim_(x->oo) f(x) exists and is finite and nonzero and if lim_(x->oo) {f(x)+(3f(x)−1)/(f^2(x))}=3 ,then the value of lim_(x->oo) f(x) is

lim_(x->oo) xsin(2/x)