Find the limit

MarkSA

Junior Member
Joined
Sep 8, 2007
Messages
243
Hi,

I have the question:
Find the limit as (x,y) goes to (0,0) of: (x^4)*(y^4)/sqrt[x^8 + y^8]

I'm having trouble figuring out how to solve this. My teacher only briefly covered this in class and said that 90-95% of these problems can be done more easily by converting to polar coordinates.. So I tried that.

lim as r goes to 0 of: [(r*costheta)^4 * (r*sintheta)^4]/sqrt[(r*costheta)^8 + (r*sintheta)^8]
Factoring the 'r' out of top and bottom, I get:
lim as r goes to 0 of: (costheta)^4 * (sintheta)^4

But here I am left with theta and no r in argument left to evaluate the limit at. I'm not sure where to go from here. Can you help?
 
Nevermind.. I think I see how the polar method works. As r->0, that theta mess goes to zero since there is no r in it. Is that correct?

What i'm having trouble with now is using the 'normal' method which the teacher mentioned has '5 steps' to it. I can't find those steps in the book... But it was something on the order of doing lim x->0 for y=0, and lim x->0 for y=x, etc. I'm not sure what the other tests are or what their significance is here... I know that for the limit to exist, all of the limits must be the same. I'm not sure why those particular tests are used though (or what all 5 are)
 
I think something went wrong. I get:

[(r*costheta)^4 * (r*sintheta)^4]
------------------------------------ =
sqrt[(r*costheta)^8 + (r*sintheta)^8]


r^8 [(cos t)^4 * (sin t)^4]
-------------------------------- =
r^4 sqrt[(cos t)^8 + (sin t)^8]


r^4 [(cos t)^4 * (sin t)^4]
---------------------------- =
sqrt[(cos t)^8 + (sin t)^8]


Now the denominator is bounded away from zero.

D[(cos t)^8 + (sin t)^8] =

8 (cos t)^7 (- sin t) + 8 (sin t)^7 cos t

8 cos t sin t [- (cos t)^6 + (sin t)^6]

4 sin 2t [- (cos t)^6 + (sin t)^6]

sin 2t = 0

2t = 0, pi, 2pi, etc.

t = 0, pi/2, pi, etc.

OR

- (cos t)^6 + (sin t)^6 = 0

(sin t)^6 = (cos t)^6

abs (sin t) = abs(cos t)

==> t = pi/4, 3pi/4, etc.

Those are your minimums, and (cos t)^8 + (sin t)^8 is nonzero at each.
 
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