Let f(x,y)={xy2x2+3y2,(x,y)≠(0,0)0,(x,y)=(0,0) Discuss continuity of f(x,y) and existence of first order partial derivateves.

Question

Let  f(x,y)={xy2x2+3y2,(x,y)≠(0,0)0,(x,y)=(0,0)  Discuss continuity of f(x,y) and existence of first order partial derivateves.

Royce Moore · Accepted Answer

Given information:  The function is f(x,y)={xy2x2+3y2,(x,y)≠(0,0)0,(x,y)=(0,0)  Calculation:  Check the continuity of the function at point (0,0).  The function is 0 at point (0,0)  Suppose the function xy2x2+3y2 approach a point (0,0) along the line y=x  Substitute y=x in the function xy2x2+3y2 and take the limit as y approaches 0 that means function approach a point (0,0) along y -axis.  lim(x,y)→(0,0)y⋅y2y2+3y2=lim(x,y)→(0,0)y25y2  =lim(x,y)→(0,0)15  =15  As we can see that lim(x,y)→(0,0)f(x,y)=15≠f(0,0). It does not satisfy the condition of continuity.  Answer: Therefore, the function is not continuous.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Let f(x,y)=\begin{cases}\frac{xy}{2x^2+3y^2},(x,y)\ne(0,0)\\0,(x,y)=(0,0)

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