Question #217526

 Consider any non-zero point in 𝑅2 and name it (π‘Ž,𝑏). Then                 

(i). Write any four different paths that passes through your chosen point (π‘Ž,𝑏).

(ii). Compute the limits of the following function when (π‘₯,𝑦) β†’ (π‘Ž,𝑏) along all these four paths,                            

 π‘“(π‘₯,𝑦) = {

(π‘₯βˆ’π‘Ž)2(π‘¦βˆ’π‘) (π‘₯βˆ’π‘Ž)4+(π‘¦βˆ’π‘)2

, (π‘₯,𝑦) β‰  (π‘Ž,𝑏) 0, (π‘₯,𝑦) = (π‘Ž,𝑏)  

(iii). Conclude from the results obtained in (ii) and answer whether lim (π‘₯,𝑦)β†’(π‘Ž,𝑏) 𝑓(π‘₯,𝑦) exists or not.  

(iv). Is the function 𝑓(π‘₯,𝑦) continuous at the origin? Explain.  

(v). Also calculate 𝑓π‘₯(π‘Ž,𝑏) and 𝑓𝑦(π‘Ž,𝑏).  

(vi). Write all points of differentiability of 𝑓.


1
Expert's answer
2021-07-19T05:52:56-0400

i.

We consider (a,b)=(0,0)(a,b) = (0,0)

Now the four different paths through (0,0)(0,0) are

y=xy=x2y=x3y=x4y=x\\ y=x^2\\ y=x^3\\ y=x^4\\


ii

f(x,y)={(xβˆ’a)2(yβˆ’b)(xβˆ’a)4(yβˆ’)2if (x,y)=ΜΈ(a,b)0if (x,y)=(a,b)But(a,b)=(0,0)f(x,y)={x2yx4y2if (x,y)=ΜΈ(0,0)0if (x,y)=(0,0)Now,lim⁑(x,y)β†’(0,0)f(x,y)=lim⁑(x,y)β†’(0,0)x2yx4y2y=xlim⁑(x)β†’(0)x2βˆ—xx4βˆ—x2=0y=x2lim⁑(x)β†’(0)x2βˆ—x2x4βˆ—x4=0.5y=x3lim⁑(x)β†’(0)x2βˆ—x3x4βˆ—x6=0y=x4lim⁑(x)β†’(0)x2βˆ—x4x4βˆ—x8=0f(x,y) = \begin{cases} \frac{(x-a)^2(y-b)}{(x-a)^4(y-)^2} &\text{if } (x,y)\not=(a,b)\\ 0 &\text{if } (x,y)=(a,b) \end{cases}\\\\ But (a,b) = (0,0)\\ f(x,y) = \begin{cases} \frac{x^2y}{x^4y^2} &\text{if } (x,y)\not=(0,0)\\ 0 &\text{if } (x,y)=(0,0) \end{cases}\\\\ Now, \lim\nolimits_{(x,y) \to (0,0)}f(x,y)= \lim\nolimits_{(x,y) \to (0,0)}\frac{x^2y}{x^4y^2}\\ y=x\\ \lim\nolimits_{(x) \to (0)}\frac{x^2*x}{x^4*x^2}=0\\ y=x^2\\ \lim\nolimits_{(x) \to (0)}\frac{x^2*x^2}{x^4*x^4}=0.5\\ y=x^3\\ \lim\nolimits_{(x) \to (0)}\frac{x^2*x^3}{x^4*x^6}=0\\ y=x^4\\ \lim\nolimits_{(x) \to (0)}\frac{x^2*x^4}{x^4*x^8}=0\\


iii.

We will conclude that along every path chosen the limit of the function exist.


iv.

The function is continuous at the origin along every path chosen except along y=x2y=x^2 7


v.

f(x,y)=x2yx4+y2fn=βˆ‚fβˆ‚x=(x2+y2)βˆ—2xyβˆ’x2y(4x3)(x4+y2)2fn(0,0)=Nonexistanefn=βˆ‚fβˆ‚y=(x4+y2)βˆ—x2βˆ’x2y(2y)(x4+y2)2fn(0,0)=Nonexistanef(x,y)=\frac{x^2y}{x^4+y^2}\\ f_n= \frac{βˆ‚f}{βˆ‚x}=\frac{(x^2+y^2)*2xy-x^2y(4x^3)}{(x^4+y^2)^2}\\ f_n(0,0)= Nonexistane\\ f_n= \frac{βˆ‚f}{βˆ‚y}=\frac{(x^4+y^2)*x^2-x^2y(2y)}{(x^4+y^2)^2}\\ f_n(0,0)= Nonexistane\\

vi.

The function is differentiable at every point except at (0,0)



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