1

I'm trying to study continuity of a function, but I can't find a way to prove it.

$$f(x,y) = \begin{cases} \frac{1-\cos(x^2y)}{x^8+y^4}&\text{for } (x,y)\ne(0,0)\\ 0 &\text{for }(x,y) =(0,0) \end{cases}$$

I am thinking of finding $2$ expressions for $(x,y)\to(0,0)$ and to apply it to the function, but I'm pretty sure that it's not correct.

Another User
  • 5,048
arcilli
  • 69

1 Answers1

0

Using the fact that when $X\to 0$,

$$1-\cos (X)=2\sin^2 (X/2)\sim \frac {X^2}{2} $$

we find that $$\lim_{x\to0}f (x,x^2)=\frac {1}{4}\ne f (0,0). $$

$f $ is not continuous at $(0,0) $.

hamam_Abdallah
  • 62,951
  • The first approximation... where is it from? – arcilli Nov 28 '17 at 21:46
  • @arcilli Don't you know that $\sin (X)\sim X $ or in other terms $\lim_0\frac {\sin (X)}{X}=1$ – hamam_Abdallah Nov 28 '17 at 21:48
  • Showing that taking the limits, as (x, y) goes to (0, 0) along two different paths, give two different answer is sufficient to show that the actual limit does not exist. However, showing that the limits along two different paths ARE the same is not sufficient to show that the limit exists. – user247327 Nov 28 '17 at 21:49
  • Sure, I know that fundamental limit and I understand how it's deducted the first relation. But, I dont figure it out how it's implemented in the second one – arcilli Nov 28 '17 at 22:01