# please enter and help me>>>thanks

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• Oct 17th 2008, 10:07 AM
flower3
please enter and help me>>>thanks
find the limits by referring details when you solve them when (x,y)goes to (0,0) :

1_ lim (x^2 + y^2 ) sin ( 1\ (x
^2 + y^2))(i'm trying but i don't know )

2- lim e ^ -1 over (sqr root for
x^2 + y^2)

3- lim
{ e ^ -1 over (sqr root for x^2 + y^2) } over (sqr root for x^2 + y^2)

4- lim 1-
( x^2 + y^2) \ ( x^2 + y^2)

5- lim y ln
( x^2 + y^2)

when i solve the limit #4 first i substitute (0,0) the answer 1\0 i know that the limit doesn't exist but i don't know how i can show that ???

in #5 when i translate xy-coordinate to r,theta i can't complete

• Oct 17th 2008, 10:14 AM
Moo
Hello,

Have you tried converting all rectangular coordinates into polar coordinates ??

$x=r \cos \theta$
$y=r \sin \theta$

$x^2+y^2=r^2$

$(x,y) \to (0,0) \Leftrightarrow r \to 0$
• Oct 17th 2008, 10:56 AM
flower3
Quote:

Originally Posted by Moo
Hello,

Have you tried converting all rectangular coordinates into polar coordinates ??

$x=r \cos \theta$
$y=r \sin \theta$

$x^2+y^2=r^2$

$(x,y) \to (0,0) \Leftrightarrow r \to 0$

yes, but i don't know why we can't use Hôpital's rule when we do parametrization for x and y ??? (we get one variable is t )

• Oct 17th 2008, 11:34 AM
ThePerfectHacker
Quote:

Originally Posted by flower3
1_ lim (x^2 + y^2 ) sin ( 1\ (x[/FONT][/FONT][FONT=Tahoma][FONT=Tahoma] ^2 + y^2))(i'm trying but i don't know )

For this one note that,
$0\leq (x^2+y^2) \sin \frac{1}{x^2+y^2} \leq (x^2+y^2)$
• Oct 19th 2008, 12:57 PM
flower3
please help me>>>>>>i don't know how
please help me>>>>>>i don't know how
• Oct 19th 2008, 01:17 PM
rman144
Reply
I'll help you on the first one because your notation is a little off on the rest:

(x^2+y^2) sin(1/(x^2+y^2))

Let x^2+y^2=r^2 through parametric transformation:

(r^2) sin(1/(r^2))

Now if we know that r=0 (because x^2+y^2=0), then we know we have:

0* sin(1/(r^2))

Regardless of what the sin value produces, you know it is within the limits of the sine function solution (-1 through +1), and because zero times any constant, even in this case zero, should the sine value equal zero, would still be zero^2=0.

Can you get it from here?