Question #4409

Show that if a,b>=0; c,d>0; with a/c + b/d > 1 then:
lim[(x,y)->(0,0)] (|x|^a*|y|^b) / (|x|^c+|y|^d) = 0
Thanks!

Expert's answer

Denote u = |x|^c& and& v = |y|^d, so

& |x|=u^(1/c)

and |y|=v^(1/d).

Then (x,y) -> (0,0) implies (u,v) -> (0,0) as

well, since c,d>0.

Therefore

& (|x|^a*|y|^b) / (|x|^c+|y|^d) =

|u|^(a/c) * |v|^(b/d) / ( u+v ).

Now use polar coordinates, so

& u =

r |cos phi|,

& v = r |sin phi|

Notice that for any phi we have that

|cos phi| + |sin phi| > cos pi/4 =

1/sqrt(2)

Hence

(|x|^a*|y|^b) / (|x|^c+|y|^d) = |u|^(a/c) * |v|^(b/d)

/ ( u + v )

= r^(a/c) |cos phi|^a * r^(b/d)

|sin phi|^b / (r^c |cos phi| + r^d |sin phi|)

= r^(a/c + b/d) |cos phi|^a |sin phi|^b / (r |cos phi| + r |sin phi|)

& lt;= r^(a/c + b/d) / r (|cos phi| + |sin

phi|)

& lt;= sqrt(2) r^(a/c + b/d) / r

& lt;= sqrt(2) r^(a/c + b/d - 1)

But

a/c + b/d > 1, hence r^(a/c + b/d - 1) -> 0 when

r->0.

lim[(x,y)->(0,0)] (|x|^a*|y|^b) / (|x|^c+|y|^d) <=

lim[(x,y)->(0,0)]& sqrt(2) r^(a/c + b/d - 1) = 0

& |x|=u^(1/c)

and |y|=v^(1/d).

Then (x,y) -> (0,0) implies (u,v) -> (0,0) as

well, since c,d>0.

Therefore

& (|x|^a*|y|^b) / (|x|^c+|y|^d) =

|u|^(a/c) * |v|^(b/d) / ( u+v ).

Now use polar coordinates, so

& u =

r |cos phi|,

& v = r |sin phi|

Notice that for any phi we have that

|cos phi| + |sin phi| > cos pi/4 =

1/sqrt(2)

Hence

(|x|^a*|y|^b) / (|x|^c+|y|^d) = |u|^(a/c) * |v|^(b/d)

/ ( u + v )

= r^(a/c) |cos phi|^a * r^(b/d)

|sin phi|^b / (r^c |cos phi| + r^d |sin phi|)

= r^(a/c + b/d) |cos phi|^a |sin phi|^b / (r |cos phi| + r |sin phi|)

& lt;= r^(a/c + b/d) / r (|cos phi| + |sin

phi|)

& lt;= sqrt(2) r^(a/c + b/d) / r

& lt;= sqrt(2) r^(a/c + b/d - 1)

But

a/c + b/d > 1, hence r^(a/c + b/d - 1) -> 0 when

r->0.

lim[(x,y)->(0,0)] (|x|^a*|y|^b) / (|x|^c+|y|^d) <=

lim[(x,y)->(0,0)]& sqrt(2) r^(a/c + b/d - 1) = 0

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