# For $a=\frac 14$, $f$ is continuous and differentiable.

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 $$2|xy | \leq x^{2}+y^{2}$$.