multivariable limit: lim cos[(x^3 - y^3)/(x^2 + y^2)]

cheffy

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Jan 10, 2007
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Find the limit of f(x,y) as (x,y) -> (0,0)

\(\displaystyle \
f(x,y) = \cos \left( {\frac{{x^3 - y^3 }}{{x^2 + y^2 }}} \right)
\\)

My intuition says that this DNE, but I don't know what path to plug in to prove it. (or my intuition is wrong) Thanks!
 
I disagree with your intuition:

Use parametric forms: x=rcos(t),  y=rsin(t),  r=x2+y3\displaystyle x = r\cos (t),\;y = r\sin (t),\;r = \sqrt {x^2 + y^3 }

Then x3y3x2+y2=r3[cos3(t)sin3(t)]r2=r[cos3(t)sin3(t)]\displaystyle \frac{{x^3 - y^3 }}{{x^2 + y^2 }} = \frac{{r^3 \left[ {\cos ^3 (t) - \sin ^3 (t)} \right]}}{{r^2 }} = r\left[ {\cos ^3 (t) - \sin ^3 (t)} \right].
Thus (x,y)(0,0)    r0\displaystyle (x,y) \to (0,0)\; \Rightarrow \;r \to 0 so x3y3x2+y2=r[cos3(t)sin3(t)]0\displaystyle \frac{{x^3 - y^3 }}{{x^2 + y^2 }} = r\left[ {\cos ^3 (t) - \sin ^3 (t)} \right] \to 0.

Now we get cos(0)=1\displaystyle \cos (0) = 1.

Is it possible that the problem is different from the one posted?
 
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