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

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The discussion revolves around evaluating the limit of the expression y*ln(y^2+x^2) as (x,y) approaches (0,0). The user attempted to use polar coordinates but encountered difficulties due to the presence of sin(θ). They explored various paths, including y=0 and x=y^2, all yielding a limit of zero. An edit suggests a potential solution using polar substitution, questioning if r*sin(θ) * ln(r^2) is less than or equal to r*ln(r^2). A response clarifies that absolute values are necessary for the inequality to hold, indicating a step towards resolving the limit.
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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 \leq r*ln r2
because if its right then problem solved :P
 
Last edited:
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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 \leq r*ln r2
because if its right then problem solved :P

Almost. You need absolute values:

|r*sinθ * ln r2| \leq r*|ln (r2)|
 
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