Question

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!

Answer 1

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