$lim_(x,y) to (0,0)fracyx = 0 Longleftrightarrow fracac + fracbd > 1$

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Let $a,b,c,d$ be positive real numbers. Show that the limit
$$lim_(x,y) to (0,0)fracyy = 0$$
if only if
$$fracac + fracbd > 1.$$




This question seems to be basically about algebraic manipulation, but I cannot seem to find a good way. Can anybody help me?




real-analysis limits




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    Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
    – Cesareo
    Jul 25 at 16:33














up vote
3
down vote

favorite
1













Let $a,b,c,d$ be positive real numbers. Show that the limit
$$lim_(x,y) to (0,0)fracyy = 0$$
if only if
$$fracac + fracbd > 1.$$




This question seems to be basically about algebraic manipulation, but I cannot seem to find a good way. Can anybody help me?




real-analysis limits




share|cite|improve this question















  • 1




    Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
    – Cesareo
    Jul 25 at 16:33












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Let $a,b,c,d$ be positive real numbers. Show that the limit
$$lim_(x,y) to (0,0)fracyy = 0$$
if only if
$$fracac + fracbd > 1.$$




This question seems to be basically about algebraic manipulation, but I cannot seem to find a good way. Can anybody help me?




real-analysis limits




share|cite|improve this question












Let $a,b,c,d$ be positive real numbers. Show that the limit
$$lim_(x,y) to (0,0)fracyy = 0$$
if only if
$$fracac + fracbd > 1.$$




This question seems to be basically about algebraic manipulation, but I cannot seem to find a good way. Can anybody help me?





real-analysis limits





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 25 at 15:48









Lucas Corrêa

1,106319




1,106319







  • 1




    Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
    – Cesareo
    Jul 25 at 16:33












  • 1




    Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
    – Cesareo
    Jul 25 at 16:33







1




1




Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
– Cesareo
Jul 25 at 16:33




Making $x=r costheta, y = rsintheta$ we have that $a+b > max(c,d)$ suffices.
– Cesareo
Jul 25 at 16:33










1 Answer
1






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Using substitutions $xto x^1/c$ and $yto x^1/d$ the problem reduce to
$$lim_(x,y) to (0,0)fracx = 0$$
iff $r+s>1$ where $r=dfracac$ and $s=dfracbd$. With polar substitutions $x=rhocostheta$ and $y=rhosintheta$ we have
$$lim_(x,y) to (0,0)fracx = lim_(rho,theta) to (0,0)frac^r = 0$$
iff $r+s>1$.






share|cite|improve this answer





















  • I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
    – Lucas Corrêa
    Jul 25 at 17:00










  • You are welcome.
    – Nosrati
    Jul 25 at 17:01










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Using substitutions $xto x^1/c$ and $yto x^1/d$ the problem reduce to
$$lim_(x,y) to (0,0)fracx = 0$$
iff $r+s>1$ where $r=dfracac$ and $s=dfracbd$. With polar substitutions $x=rhocostheta$ and $y=rhosintheta$ we have
$$lim_(x,y) to (0,0)fracx = lim_(rho,theta) to (0,0)frac^r = 0$$
iff $r+s>1$.






share|cite|improve this answer





















  • I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
    – Lucas Corrêa
    Jul 25 at 17:00










  • You are welcome.
    – Nosrati
    Jul 25 at 17:01














up vote
1
down vote



accepted










Using substitutions $xto x^1/c$ and $yto x^1/d$ the problem reduce to
$$lim_(x,y) to (0,0)fracx = 0$$
iff $r+s>1$ where $r=dfracac$ and $s=dfracbd$. With polar substitutions $x=rhocostheta$ and $y=rhosintheta$ we have
$$lim_(x,y) to (0,0)fracx = lim_(rho,theta) to (0,0)frac^r = 0$$
iff $r+s>1$.






share|cite|improve this answer





















  • I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
    – Lucas Corrêa
    Jul 25 at 17:00










  • You are welcome.
    – Nosrati
    Jul 25 at 17:01












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Using substitutions $xto x^1/c$ and $yto x^1/d$ the problem reduce to
$$lim_(x,y) to (0,0)fracx = 0$$
iff $r+s>1$ where $r=dfracac$ and $s=dfracbd$. With polar substitutions $x=rhocostheta$ and $y=rhosintheta$ we have
$$lim_(x,y) to (0,0)fracx = lim_(rho,theta) to (0,0)frac^r = 0$$
iff $r+s>1$.






share|cite|improve this answer













Using substitutions $xto x^1/c$ and $yto x^1/d$ the problem reduce to
$$lim_(x,y) to (0,0)fracx = 0$$
iff $r+s>1$ where $r=dfracac$ and $s=dfracbd$. With polar substitutions $x=rhocostheta$ and $y=rhosintheta$ we have
$$lim_(x,y) to (0,0)fracx = lim_(rho,theta) to (0,0)frac^r = 0$$
iff $r+s>1$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 25 at 16:47









Nosrati

19.3k41544




19.3k41544











  • I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
    – Lucas Corrêa
    Jul 25 at 17:00










  • You are welcome.
    – Nosrati
    Jul 25 at 17:01
















  • I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
    – Lucas Corrêa
    Jul 25 at 17:00










  • You are welcome.
    – Nosrati
    Jul 25 at 17:01















I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
– Lucas Corrêa
Jul 25 at 17:00




I tried using polar coordinates, but I didn't think of the substitution you did. Thank you!
– Lucas Corrêa
Jul 25 at 17:00












You are welcome.
– Nosrati
Jul 25 at 17:01




You are welcome.
– Nosrati
Jul 25 at 17:01












 

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