Vtyjkyuk

suppose $f(x,y)=frac^bx$ and $a,b,c,d >0$ how would you show that if $fracac + fracbd >1$ then $limlimits_(x,y)to(0,0)frac^bx$ exists and equals $0$?

I've been trying to use the squeeze theorem and set up an inequality but im really struggling.

Young's inequality states that for $x,y>0$ and $p,q>0$ with $frac1p+frac1q=1$ we have
$xyle fracx^pp+fracy^qq$ so $|x|^a|y|^ble frac1p|x|^ap+frac1q|y|^bq$. We can choose $p$ and $q$ such that $ap>c$ and $bq>d$:

Assume for a contradiction that for all $p,q$ with $frac1p+frac1q=1$ we have $aple c$ and $bqle d$ then $frac a c +frac bd le frac1p+frac 1q=1$.

Now
$$frac^bx le frac^app + fracq le |x|^ap-c+|y|^bq-d,
$$
which tends to $0$ as $(x,y)to (0,0)$.

The appropriate general technique is Lagrange multipliers. Because of the absolute value signs, we may ask for the largest value, with both $x,y > 0,$ of
$$ fracx^a y^bx^c + y^d $$
The multiplier condition is that the ratio $a x^a-1 y^b ::: c x^c-1$ is the same as $b x^a y^b-1 ::: d y^d-1.$ Using cross-multiplication for fractions that are supposed to be equal gives
$$ adx^a-1 y^b+d-1 = bc x^a+c-1 y^b-1. $$ The first cancellation gives
$$ ad y^b+d-1 = bc x^c y^b-1. $$ The next gives
$$ ad y^d = bc x^c . $$
That is, the largest value for fixed $x^c + y^d$ occurs when
$$ y = lambda x^c/d $$
and $lambda$ is a positive constant.
$$ fracx^a y^bx^c + y^d leq fraclambda_2 x^a x^fracbcdx^c + lambda_3 x^c leq lambda_4 x^a x^fracbcd x^-c$$
The resulting limit as $x$ goes to $0$ is also $0$ when
$$ a + fracbcd - c > 0 ; , ; $$
$$ a + fracbcd > c ; , ; $$
or, as $c>0,$
$$ fracac + fracbd > 1 ; . ; $$

Let consider

• $|x|=|u|^1/c$


• $|y|=|v|^1/d$

then we have that

$$frac^bx=frac=r^frac a c+frac b d-1cdot f(theta)to 0$$

since $f(theta)=frac^b/dsin theta$ is bounded and for $frac a c+frac b d>1 iff frac a c+frac b d-1>0$ we have that $r^frac a c+frac b d-1to 0$.