`Lim_(x -> 0) (1/x)^2sinx​`

Evaluate : lim_( x -> 0 ) ( 1/sinx - 1/x )

Evaluate : lim_( x -> 0 ) ( (e^N - sinx - 1) )/a

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

lim_(x to 0) (x^3 sin(1/x))/sinx

lim_(x to 0) (e^x-e^sinx)/(x-sinx)

The value of the limit lim_(x to 0) ((x)/(sinx))^(6//x^(2)) is

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

Evaluate the following : lim_(x to 0)(x^(2))/(sinx^(2))

The value of lim_(x to 0) (e^x-e^sinx)/(2(x-sinx)) , is