`g(t)=lim_(x->0)(1+atanx)^(1//x)`, a is even prime number, then `g(2)=`

g(x),=x^(c)e^(cx) and f(x)=int_(0)^(x)te^(2t)(1+3t^(2))^((1)/(2))dt* if L,=lim_(x rarr oo)(f'(x))/(g'(x)) is non-zero finit number then :,

Let g (x ) = x ^(C )e ^(Cx) and f (x) = int _(0)^(x) te ^(2t) (1+3t ^(2))^(1//2) dt. If L = lim _(x to oo) (f'(x ))/(g '(x)) is non-zero finite number then : The value of L . Is :

Let g (x ) = x ^(C )e ^(Cx) and f (x) = int _(0)^(x) te ^(2t) (1+3t ^(2))^(1//2) dt. If L = lim _(x to oo) (f'(x ))/(g '(x)) is non-zero finite number then : The value of L . Is :

Statement-1:lim_(x rarr0)((g(2-x^(2)))/(g(2)))^((4)/(x^(2)))=e^(-(1)/(2)) where g(2)=-40 and g(2)=-5 Statement-2: If lim_(x rarr a)f(x)=1,lim_(x rarr a)g(x)rarr oo then lim_(x rarr a){f(x)}^(g(x))=e^(lim_(a))(f(x)-1)g(x)

Given that for each a in (0,1)lim_(x to 0) int_(h)^(-h) t^(-a)(1-t)^(a-1)dt exits and is equal to g(a).If g(a) is differentiable in (0,1), then the value of g'((1)/(2)) , is

Let g(x)=((x-1)^(n))/(logcos^(m)(x-1)),0ltxlt2 m and n integers, m ne0, n gt0 and. If lim_(xrarr1+) g(x)=-1 , then

If lim_(xrarr0) (f(x))/(x^(2))=a and lim_(xrarr0) (f(1-cosx))/(g(x)sin^(2)x)=b (where b ne 0 ), then lim_(xrarr0) (g(1-cos2x))/(x^(4)) is

If lim_(xrarr0) (f(x))/(x^(2))=a and lim_(xrarr0) (f(1-cosx))/(g(x)sin^(2)x)=b (where b ne 0 ), then lim_(xrarr0) (g(1-cos2x))/(x^(4)) is

If lim_(x rarr a)[f(x)+g(x)]=2 and lim_(x rarr a)[f(x)-g(x)]=1, then find the value of lim_(x rarr a)f(x)g(x)