`lim_(x->0)xlog(sinx)`

lim_(x->0) xlog x

lim_(x->0)|sinx|/x

lim_(x->0) ([(-5sinx)/x]+[(6sinx)/x] .(where [-] denotes greatest integer function) is equal to

if l=lim_(x->0) (x(1+acosx) - bsinx)/x^3 = lim_(x->0) (1+acosx)/x^2-lim_(x->0) (b sinx)/x^3 where l in R , then

if l=lim_(x->0) (x(1+acosx) - bsinx)/x^3 = lim_(x->0) (1+acosx)/x^2-lim_(x->0) (b sinx)/x^3 where l in R , then

Evaluate lim_(xto0)|x|^(sinx)

What is the value of lim_(xto0) (sinx)/(x) ?

If a=lim_(x->oo) sinx/x & b=lim_(x->0) sinx/x Then int (ab log(1+x)+x^2)dx is equal to

Evaluate : lim_( x -> 0 ) ( ( sinx - x ) /x^2 )