Let $f(x,y) = \begin{cases} \frac{xy}{(x^2+y^2)^a}, & \text{if $(x,y)\neq (0,0)$ } \\[2ex] 0, & \text{if $(x,y)= (0,0)$} \end{cases}$

Then which one of the following is TRUE for $f$ at the point $(0,0)$?

(A) For $a=1$, $f$ is continuous but not differentiable.

(B) For $a=\frac 12$, $f$ is continuous and differentiable.

(C) For $a=\frac 14$, $f$ is continuous and differentiable.

(D) For $a=\frac 34$, $f$ is neither continuous nor differentiable

My attempt:

(A) For $a=1$ $f(x,y) = \begin{cases} \frac{xy}{(x^2+y^2)}, & \text{if $(x,y)\neq (0,0)$ } \\[2ex] 0, & \text{if $(x,y)= (0,0)$} \end{cases}$

is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) = \begin{cases} \frac{m}{(1+m^2)}, & \text{if $x\neq 0$ } \\[2ex] 0, & \text{if $x= 0$} \end{cases}$

not continuous at $0$. So, (A) is not true.

(B) For $a=\frac 12$ $f(x,y) = \begin{cases} \frac{xy}{(x^2+y^2)^{\frac 12}}, & \text{if $(x,y)\neq (0,0)$ } \\[2ex] 0, & \text{if $(x,y)= (0,0)$} \end{cases}$

is not differentiable at $(0,0)$ as $\frac {f(h,k)-f(0,0)}{\sqrt{h^2+k^2}}= \frac{hk}{(h^2+k^2)}$ where $(h,k) \to (0,0)$ if we put $k=mh$ then

$f(h,mh) =\frac{m}{(1+m^2)} \neq (0,0) $

Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.

(C) For $a=\frac 14$ $f(x,y) = \begin{cases} \frac{xy}{(x^2+y^2)^{\frac 14}}, & \text{if $(x,y)\neq (0,0)$ } \\[2ex] 0, & \text{if $(x,y)= (0,0)$} \end{cases}$

Then, $f(x,y) =\frac{x\sqrt y}{(\frac{x^2}{y^2}+1)^{\frac 14}}$ Can we prove continuity of $f$ from here?

Observe, $f$ is differentiable at $(0,0)$ as $\frac {f(h,k)-f(0,0)}{\sqrt{h^2+k^2}}= \frac{hk}{(h^2+k^2)^\frac 34}=\frac{h}{(\frac{h^2}{k^{4/3}}+k^{2/3})^{\frac 34}} \to (0,0)$ where $(h,k) \to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.

(D) For $a=\frac 34$ $f(x,y) = \begin{cases} \frac{xy}{(x^2+y^2)^{\frac 34}}, & \text{if $(x,y)\neq (0,0)$ } \\[2ex] 0, & \text{if $(x,y)= (0,0)$} \end{cases}$

From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.

So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.


Recall that $$|xy|\leq \frac{(x^2+y^2)}{2}$$ and so $$\frac{|xy|}{(x^2+y^2)^{1/4}}\leq \frac{(x^2+y^2)^{3/4}}{2}\to 0$$ as $(x,y)\to (0,0).$ Hence you have continuity at $(0,0).$


For $a=\frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ \frac {xy} {(x^{2}+y^{2})^{3/4}} \to 0$ which follows easily from the fact that $2|xy | \leq x^{2}+y^{2}$.


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