Whether the options $(iii)$ and $(iv)$ make sense?

65 Views Asked by At

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.

1

There are 1 best solutions below

9
On BEST ANSWER

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.