If I write the problem, it will be like this:Lim (y->0) {Lim (x->0) (x sin y - y sin x) / (x2 + y2)}.And calculate the FIRST limit we getLim (y->0) 0 = 0.Since x and y both approach 0, we can take the sin(u)=u-u^3/6 approximation as close enough for small u. Then(x*sin(y)-y*sin(x))/(x^2+y^2)=(x*(y-y^3/6)-y*(x-x^3/6))/(x^2+y^2)=(x*y^3-x^3*y)/(6*(x^2+y^2))=x*y*(y^2-x^2)/(6*(x^2+y^2)) Now we can see how this behaves by noting the 3 cases y>>x, y<0x*y*(-x^2)/(6*x^2)=-x*y/6 which is 0 as x,y->0 andx*t*x*(t^2*x^2-x^2)/(6*(x^2+t^2*x^2))=x^2*t*(t^2–1)/(6*(t^2+1)) and not matter what REAL value t has, this is 0 as x (and HENCE y)->0The denominator is non-negative, so there is no path y=y(x) which could make the denom vanish, unlike the problemx*y/(x^2-y^2) which along the line y=m*x has a value m/(1-m^2) but along the curve y=x^2 we get x^3/(x^2-x^4)=x/(1-x^2) which goes to 0 as x->0.Step-by-step explanation:pls follow thank and vote me pls pls pls pls pls pls pls pls pls pls pls