Does y*ln(y^2+x^2) Approach Zero as (x, y) Goes to (0,0)?

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SUMMARY

The limit of the expression y*ln(y^2+x^2) as (x, y) approaches (0, 0) is confirmed to approach zero. The discussion highlights the use of polar coordinates to simplify the evaluation, specifically substituting r and θ, where r represents the distance to the origin. The conclusion is reached by demonstrating that |r*sinθ * ln(r^2)| is bounded by r*|ln(r^2)|, leading to the limit being zero as r approaches zero.

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Homework Statement



(x,y)-->(0,0) y*ln (y2+x2)

Homework Equations



!

The Attempt at a Solution


i've tried polar coordinates i end up with sinθ in it don't know how to proceed

and I've taken several paths like y=0 , x=0 , x=y2 , x=y , x=y3
they all yeild zero

Edit : is it right if i after making polar sub. to say that :

r*sinθ * ln r2 [itex]\leq[/itex] r*ln r2
because if its right then problem solved :P
 
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AHMAD MS said:

Homework Statement



(x,y)-->(0,0) y*ln (y2+x2)

Homework Equations



!

The Attempt at a Solution


i've tried polar coordinates i end up with sinθ in it don't know how to proceed

and I've taken several paths like y=0 , x=0 , x=y2 , x=y , x=y3
they all yeild zero

Edit : is it right if i after making polar sub. to say that :

r*sinθ * ln r2 [itex]\leq[/itex] r*ln r2
because if its right then problem solved :P

Almost. You need absolute values:

|r*sinθ * ln r2| [itex]\leq[/itex] r*|ln (r2)|
 

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