1

Let$~f:\mathbb{R}^2\longrightarrow \mathbb{R}$ be such that $f_x=\dfrac{x}{\sqrt{x^2-y^2}}$ and $f_y=\dfrac{y}{\sqrt{x^2-y^2}}$,$x^2\neq y^2$. Consider the following statements $:$

$(i)$ $\displaystyle \lim_{{(x,y)\to (2,-1) }} f(x,y)$ exists.

$(ii)$ $f(x,y)$ is continuous at $(2,-1)$.

$(iii)$ $f_x$ and $f_y$ are not continuous at $(0,0)$.

$(iv)$ $f_x$ and $f_y$ are continuous everywhere.

then which of the above statement/s is/are true?

Clearly $(0,0)$ is not in the domain of $f_x$ and $f_y$. So there is no question of continuity or discontinuity there. So $(iii)$ and $(iv)$ are absurd options. At $(2,-1)$ both $f_x$ and $f_y$ are continuous so, $f$ is differentiable at $(2,-1)$ and hence $f$ is continuous at $(2,-1)$. Therefore according to me $(ii)$ is the only correct option above.

Is this reasoning correct at all? Please check it.

Thank you in advance.

Arnab Chattopadhyay.
  • 2,570

1 Answers1

1

Your reasoning is almost correct.

$(iii)$ is true (it's the negation).

$(i)$ is true too because $(ii)$ is true.

Also note that for differentiability, you need the partial derivatives to exist in a neighbourhood of said point, which is pretty much obvious.

I assumed here that the domain of the partial derivatives are the same as $f$, so only "half" of their values are given in the question. I chose this, because it would be lame to give a secret function $f$ defined on $\mathbb{R}^2$ and the given partial derivatives with another domain.

The Phenotype
  • 5,199
  • You are right @The Phenotype. – Arnab Chattopadhyay. Feb 04 '18 at 17:59
  • But can we discuss about $(iii)$ because $(0,0)$ is not in the domain of $f$. – Arnab Chattopadhyay. Feb 04 '18 at 18:04
  • Even when we say a function is continuous everywhere we mean to say that the function is continuous everywhere in it's domain. To discuss continuity of a function at some point which is not in the domain of it doesn't make much sense.Isn't it? – Arnab Chattopadhyay. Feb 04 '18 at 18:07
  • 1
    @ArnabChatterjee. You're right. Here is discussed that it's a bad question. I would say that continuity implies existence, while by definition existence should be given first before talking about continuity. – The Phenotype Feb 04 '18 at 18:12
  • So the correct options are only $(i)$ and $(ii)$. Right? – Arnab Chattopadhyay. Feb 04 '18 at 18:15
  • 1
    @ArnabChatterjee. You could say that $(iii)$ is undefined :) – The Phenotype Feb 04 '18 at 18:16
  • Then $(iv)$ is also true. – Arnab Chattopadhyay. Feb 04 '18 at 18:23
  • 1
    @ArnabChatterjee. Actually I see now that the partial derivatives here are given for $x^2-y^2> 0$, which doesn't mean that it's not defined for the other values. It could be that for the other values, the partial derivatives are equal to $0$, as partial derivatives in general don't have to be continuous. So my answer holds in the case that they are defined on $\mathbb{R}^2$, as the limit (infinity or something) will not be equal to the (not given here) value in the point. If the domain is only $x^2-y^2> 0$, then $(iv)$ is indeed true and $(iii)$ undefined. Good catch! – The Phenotype Feb 04 '18 at 21:01
  • Thanks @The Phenotype for your valuable suggestions. – Arnab Chattopadhyay. Feb 05 '18 at 08:31