# Limit problem??

• January 21st 2010, 11:30 AM
Nappy
Limit problem??
How do I evaluate this without getting zero on the denominator making it undefined??

Question is..

limit x2/sqrt(x2 + y2) = 0
(x,y) => (0,0)

x2=x squared
sqrt= square root

Also Im a newbie, how can I use proper maths symbols here??

Cheers

Nappy
• January 21st 2010, 11:35 AM
pickslides
Have you tried using polar co-ordinates?

$
x= r\cos(\theta), y= r\sin(\theta)
$
• January 21st 2010, 11:36 AM
General
Quote:

Originally Posted by Nappy
How do I evaluate this without getting zero on the denominator making it undefined??

Question is..

limit x2/sqrt(x2 + y2) = 0
(x,y) => (0,0)

x2=x squared
sqrt= square root

Also Im a newbie, how can I use proper maths symbols here??

Cheers

Nappy

Polar coordinates is good choice here.
Or, use the 2-path rule on the two paths: y=x and y=0.
• January 21st 2010, 11:41 AM
pickslides
Quote:

Originally Posted by General
Or, use the 2-path rule on the two paths: y=x and y=0.

Not sure I have seen this method, could you explain further?
• January 21st 2010, 11:44 AM
chisigma
In such cases the following change of variables ever works...

$x= \rho\cdot \cos \theta$

$y= \rho\cdot \sin \theta$ (1)

Using (1) we obtain...

$\lim_{(x,y) \rightarrow (0,0)} \frac{x^{2}}{\sqrt{x^{2}+y^{2}}}= \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \cos^{2} \theta}{\rho}= \lim_{\rho \rightarrow 0} \rho \cdot \cos^{2} \theta$ (2)

Kind regards

$\chi$ $\sigma$
• January 21st 2010, 11:51 AM
Nappy
Thanks guys,

If I was asked to prove the above directly from the definition of a limit would that answer be sufficient???