# Prove that $f$ is discontinuous at $(0,0)$

Let $f$ be defined by

$$f(x,y) = \begin{cases} \biggl\lvert \frac{y}{x^2} \biggr\rvert e^{-\bigl\lvert \frac{y}{x^2} \bigr\rvert} , \quad \text{ if x \neq 0} \\ 0, \qquad \qquad \quad \text{if x = 0.} \end{cases}$$

Prove that $f$ is discontinuous at $(0,0)$.

Prove that $f$ is continuous along any line passing through the origin.

Hi so I am really stuck here. I am not sure what method to use.

-

Hint: You could go like this. If it was continuous, then the limit of the function when $(x,y)\rightarrow(0,0)$ should be the same while approaching the point through any possible trajectory. If we then find two trajectories for which the limit is different, then it will be discontinuous: try parabolas ($y=mx^2$) and lines ($y=mx$).

For proving they're continuous on lines you will have to substitute $y=mx$ and make the same calculation as before.

Particularly, in the case of parabolas all variables vanish, and you have a constant function $|m|e^{-|m|}$, which if obviously different from $f(0,0)=(0,0)$ if $m\not=0$

-
Thank you for the hint :) –  Dafty Mar 1 '13 at 19:50

Hint:

Show that along the curve $y=x^2, \ \displaystyle \lim_{(x,y)\to(0,0)}f(x,y)\neq f(0,0).$ What is the limit along the lines $y=kx$?

-