Continuity and differentiability

**jollysa87** · July 30th 2009, 10:27 AM

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...

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

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

**Danny** · July 30th 2009, 11:50 AM

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...

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

The function is continuous when $x$ != $y^3$ but it is not when $x=y^3$ because the limit for $(x,y)->(y^3,y)$ 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 $x = m y^3, m \ne 1$ then

$<br /> \lim_{(x,y)->(0,0)} \frac{m y^3}{m y^3-y^3} = \frac{m}{m-1}<br />$ which clear changes as we very $m$ . Since we get different limits f is not continuous at $(0,0)$ .

**jollysa87** · July 30th 2009, 12:13 PM

So it's not continuous in (0,0) but what about the other points that satisfies $x=y^3$ ? Thanks

**Danny** · July 30th 2009, 12:43 PM

Originally Posted by jollysa87

So it's not continuous in (0,0) but what about the other points that satisfies $x=y^3$ ? Thanks

Still not. You have $f$ defined as 0 if $x = y^3$ . Thus, in a neighborhood of this curve, if continuous, then $f$ can be made arbitrary close to $0$ . However, along $x = my^3,$ the function is $\frac{m}{1-m}$ which is large near $m = 1$ , not zero.

**jollysa87** · July 30th 2009, 12:59 PM

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

**Danny** · July 30th 2009, 01:10 PM

Originally Posted by jollysa87

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

Yep.

**jollysa87** · July 30th 2009, 01:52 PM

So the following function too is not continuous for $y=x$ ?

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

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

**Danny** · July 30th 2009, 02:46 PM

Originally Posted by jollysa87

So the following function too is not continuous for $y=x$ ?

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

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

Yep - you got it!

Now consider the following

$<br /> \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+y^2}<br />$

Every path you follow you get 0. So, now what?

**jollysa87** · July 30th 2009, 10:16 PM

That one is continuous because using polar coordinates results 0 no matter what, am I right?

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