2 Variables function limit problem

Question

How to solve the limit problem

$$\lim_{(x,y) \to (0,0)} \frac{x^{1/5}y^{4/5}}{\sqrt{x^2+y^2}}$$

I know it doesn't exist but I want to see the way.

Thank you.

Answer 1

Hint: take the polar coordinates. We have $$\frac{\rho^{1/5}\sin^{1/5}\left(\theta\right)\rho^{4/5}\cos^{4/5}\left(\theta\right)}{\rho}=\sin^{1/5}\left(\theta\right)\cos^{4/5}\left(\theta\right).$$ Can you conclude?

Answer 2

The function $$f(x,y) = \frac{x^{1/5}y^{4/5}}{\sqrt{x^2+y^2}}$$ is not continous at $(0,0)$.

To see that, take the limit $(x,y) \to (0,0)$ in two different ways: Approach it on the line $(t,0)$ and do $t \to 0$ or do $(x,y) = (t,t)$ and do $t\to0$.

This is valid since both $(t,0)\to (0,0)$ as $t\to0$ and $(t,t)\to(0,0)$ as $t\to0$.

First approach:

$$\lim_{t \to 0} f(t,0) = \lim_{t\to0} \frac{t^{1/5}\cdot 0}{\sqrt{t^2 + 0^2}} = 0$$ but

$$\lim_{t \to 0^+} f(t,t) = \lim_{t\to0^+} \frac{t^{1/5}\cdot t^{4/5}}{\sqrt{2t^2}} = \lim_{t \to 0^+} \frac{t^{1/5 + 4/5}}{\sqrt{2}t} =\lim_{t \to 0^+} \frac{t}{\sqrt{2}t} =\lim_{t \to 0^+ } \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \neq 0 $$

So, we just found out that if we approach $(0,0)$ in two different ways, we get completly different results. Therefore, the function is not continous at $(0,0)$, i.e. the limit $\lim_{(x,y)\to(0,0)} f(x,y)$ doesn't exist.