Limit problem??

**Nappy** · January 21st 2010, 11:30 AM

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

**pickslides** · January 21st 2010, 11:35 AM

Have you tried using polar co-ordinates?

$<br /> x= r\cos(\theta), y= r\sin(\theta)<br />$

**General** · January 21st 2010, 11:36 AM

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.

**pickslides** · January 21st 2010, 11:41 AM

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?

**chisigma** · January 21st 2010, 11:44 AM

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

$\chi$ $\sigma$

**Nappy** · January 21st 2010, 11:51 AM

Thanks guys,

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

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