# Continuity and differentiability

• Jul 30th 2009, 10:27 AM
jollysa87
Continuity and differentiability
Hi all,

I got some problems doing some exams exercises...
They ask me to study the continuity and the differentiability of some functions but I always get similar results... (Doh)

$\displaystyle f(x,y)=x/(x-y^3)$, if $\displaystyle x$ != $\displaystyle y^3$
$\displaystyle f(x,y)=0$, if $\displaystyle x=y^3$

The function is continuous when $\displaystyle x$ != $\displaystyle y^3$ but it is not when $\displaystyle x=y^3$ because the limit for $\displaystyle (x,y)->(y^3,y)$ results +∞ or -∞ depending on the approaching direction... Am I wrong? If yes, how do I have to solve that limit? Thanks to all!
• Jul 30th 2009, 11:50 AM
Jester
Quote:

Originally Posted by jollysa87
Hi all,

I got some problems doing some exams exercises...
They ask me to study the continuity and the differentiability of some functions but I always get similar results... (Doh)

$\displaystyle f(x,y)=x/(x-y^3)$, if $\displaystyle x$ != $\displaystyle y^3$
$\displaystyle f(x,y)=0$, if $\displaystyle x=y^3$

The function is continuous when $\displaystyle x$ != $\displaystyle y^3$ but it is not when $\displaystyle x=y^3$ because the limit for $\displaystyle (x,y)->(y^3,y)$ results +∞ or -∞ depending on the approaching direction... Am I wrong? If yes, how do I have to solve that limit? Thanks to all!

Actually, f is not continuous. If we follow the path $\displaystyle x = m y^3, m \ne 1$then

$\displaystyle \lim_{(x,y)->(0,0)} \frac{m y^3}{m y^3-y^3} = \frac{m}{m-1}$ which clear changes as we very $\displaystyle m$. Since we get different limits f is not continuous at $\displaystyle (0,0)$.
• Jul 30th 2009, 12:13 PM
jollysa87
So it's not continuous in (0,0) but what about the other points that satisfies $\displaystyle x=y^3$? Thanks(Bow)
• Jul 30th 2009, 12:43 PM
Jester
Quote:

Originally Posted by jollysa87
So it's not continuous in (0,0) but what about the other points that satisfies $\displaystyle x=y^3$? Thanks(Bow)

Still not. You have $\displaystyle f$defined as 0 if $\displaystyle x = y^3$. Thus, in a neighborhood of this curve, if continuous, then $\displaystyle f$ can be made arbitrary close to $\displaystyle 0$. However, along $\displaystyle x = my^3,$ the function is $\displaystyle \frac{m}{1-m}$ which is large near $\displaystyle m = 1$, not zero.
• Jul 30th 2009, 12:59 PM
jollysa87
Thanks, so I can conclude that the function is not differentiable for $\displaystyle x=y^3$ because it's not continuous, right?
• Jul 30th 2009, 01:10 PM
Jester
Quote:

Originally Posted by jollysa87
Thanks, so I can conclude that the function is not differentiable for $\displaystyle x=y^3$ because it's not continuous, right?

Yep.
• Jul 30th 2009, 01:52 PM
jollysa87
So the following function too is not continuous for $\displaystyle y=x$?

$\displaystyle f(x,y)=y(1+x)/(x-y)$, if y!=x
$\displaystyle f(x,y)=0$, if y=x

Because if I follow the path $\displaystyle y=mx$, m!=1 then the limit change with $\displaystyle m$?
• Jul 30th 2009, 02:46 PM
Jester
Quote:

Originally Posted by jollysa87
So the following function too is not continuous for $\displaystyle y=x$?

$\displaystyle f(x,y)=y(1+x)/(x-y)$, if y!=x
$\displaystyle f(x,y)=0$, if y=x

Because if I follow the path $\displaystyle y=mx$, m!=1 then the limit change with $\displaystyle m$?

Yep - you got it!

Now consider the following

$\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+y^2}$

Every path you follow you get 0. So, now what?
• Jul 30th 2009, 10:16 PM
jollysa87
That one is continuous because using polar coordinates results 0 no matter what, am I right?