0

Let f: $R_2$ $→\mathbb R,φ: \mathbb R→R_2, \, ψ: \mathbb R→ \mathbb R2$ be given by φ(t)=(t,t),ψ(t)=($t^2,t$),t \, t ∈ $\mathbb R$ and

$$f(x,y) = \begin{matrix} \frac{2xy^2}{x^2+y^4} \quad \text{if (x,y)} \neq (0,0) \\ K \qquad \, \, \text{if (x,y)} = (0,0)\end{matrix} $$

"Is f continuous at $(0,0)$ for any choice of K? Explain fully."

I will say say....

Along the the path $$ φ(t)=(t,t) $$ $$\lim_{t \to 0} f(φ(t)) = 0$$

However $$ ψ(t)=(t^2,t) $$ $$\lim_{t \to 0} f(ψ(t)) = 1$$

Therefore f is not continuous a (0,0) as at least two paths differ in values.

Empty
  • 13,012
kingportable
  • 581

1 Answers1

2

NOT.

$$\lim_{(x,y)\to (0,0)}f(x,y)$$does not exist..So , the question of continuity does not arise.

Empty
  • 13,012
  • Isn't the limit K as $f(x,y) = K , , \text{if (x,y)} = (0,0)$? – kingportable Sep 01 '15 at 06:10
  • I can't understand what you are saying ... – Empty Sep 01 '15 at 06:13
  • Isn't defined that if (x,y)=(0,0) f(x,y)=K. Meaning the value is constant along any path going through (0,0) – kingportable Sep 01 '15 at 06:20
  • @Panja so all i have to do is show that it is no continuous along y=x and y^2=x as x approaches zero. Where it its limits are 0 and 1.,,,, Then say, since f is not continuous it is hence not continuous for any value of K. – kingportable Sep 01 '15 at 06:36
  • @ kingportable) That does not depend on the values of $K$. If the limit does not depend on the path (whatever the value of $K$) , that is the limit is same along any path then the limit exists.. – Empty Sep 01 '15 at 08:00
  • @Spanja So i say continuity is not dependent on K. Because continuity only occurs if the limit exists along any path for which it does not satisfy. – kingportable Sep 01 '15 at 09:06