Limit of a Multivariable functionNEED HELP

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The limit of the multivariable function [6*(x^3)*(y^2)] / [2*(x^4) + (y^4)] as (x,y) approaches (0,0) was analyzed using polar coordinates. An initial attempt incorrectly calculated the power of r in the numerator. After correcting the algebra, the limit simplifies to [6r*(cos(theta))^3*(sin(theta))^2] / [2*(cos(theta))^4 + (sin(theta))^4]. As r approaches zero, the expression approaches zero divided by a non-zero number, confirming that the limit is indeed zero. Thus, the limit of the function is zero.
*Helix*
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Homework Statement



Asked to find the limit of [6*(x^3)*(y^2)] / [2*(x^4) + (y^4)] as (x,y) is approaching (0,0)

Homework Equations



lim, as (x,y) ---> (0,0), of [6*(x^3)*(y^2)] / [2*(x^4) + (y^4)]

x = rcos(theta); y = rsin(theta)

The Attempt at a Solution



Tried numerous times with polar equations x = rcos(theta); y = rsin(theta)

came up with lim r -->0 [6r^2( (costheta)^3(sintheta)^2))] / [r^4(2*(costheta)^4 + (sintheta)^4]

D.N.E? ...I think its zero... can't prove it though...HELP!
 
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Welcome to PF, *Helix*.


*Helix* said:
came up with lim r -->0 [6r^2( (costheta)^3(sintheta)^2))] / [r^4(2*(costheta)^4 + (sintheta)^4]


There's your algebra error. The numerator has the wrong power of r.
 
ohh..so the function becomes : [6r(costheta)^3(sintheta)^2] /[2(costheta)^4 + (sintheta)^4] and as r is approaching zero ---> 0/(a number that will never be zero) is zero
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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