Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is clear that the function $\psi(x,y)=yx^{-1}$ is differentiable on $R^2-{0}$. But what about for $x=0$ -- what if I set $\psi(0,y)\equiv0$? Then how can I determine whether $\psi$ is differentiable there? Thanks.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

The new function is not continuous at any point of the form $(0,a)$; so it cannot be differentiable.

Approaching $(0,a)$, $a\neq 0$, along $x=0$ gives a limit of $0$. Approaching along any other line, $y = kx + a$ as $x\to 0$, gives $$\begin{align*} \lim_{x\to 0^+}\psi(x,kx+a) &= \lim_{x\to 0^+}\frac{kx+a}{x} = \left\{\begin{array}{ll} \infty & \text{if }a\gt 0\\ -\infty & \text{if }a\lt 0 \end{array}\right.\\ \lim_{x\to 0^-}\psi(x,kx+a) &= \lim_{x\to 0^-}\frac{kx+a}{x} = \left\{\begin{array}{ll} -\infty & \text{if }a\gt 0\\ +\infty & \text{if }a\lt 0 \end{array}\right. \end{align*}$$

At $(0,0)$, approaching along $x=ky$ gives a limit of $k$, so again the limit does not exist.

Since the function is not continuous at any point $(0,a)$, it's not differentiable either.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.