About limit of variable x y
$begingroup$
Suppose $a>0,b>0,r=sqrt{x^2+y^2}$
1)proof $x^ay^b=o(r^l)(rrightarrow 0),0<l<a+b$
$2)lim_{(x,y)rightarrow (0,0)}frac{x^ay^b}{r^{a+b}} $exist or not
My attempt
1)$$frac{x^ay^b}{r^l}=frac{x^ay^b}{(x^2+y^2)^frac{l}{2}}le frac{1}{2^l}frac{x^ay^b}{(xy)^{frac{l}{2}}}$$
And I don’t know what to do next
2)let $y=x$, $xrightarrow 0$ then I get the limit is $$frac{1}{2^{frac{a+b}{2}}}$$
and I let $y=0,xrightarrow 0$,I get the limit equal $0$ ,am I right?
Thanks for your help!
real-analysis calculus functional-analysis limits analysis
asked Jan 8 at 6:02
jacksonjackson
1379
$endgroup$
add a comment |
$begingroup$
Suppose $a>0,b>0,r=sqrt{x^2+y^2}$
1)proof $x^ay^b=o(r^l)(rrightarrow 0),0<l<a+b$
$2)lim_{(x,y)rightarrow (0,0)}frac{x^ay^b}{r^{a+b}} $exist or not
My attempt
1)$$frac{x^ay^b}{r^l}=frac{x^ay^b}{(x^2+y^2)^frac{l}{2}}le frac{1}{2^l}frac{x^ay^b}{(xy)^{frac{l}{2}}}$$
And I don’t know what to do next
2)let $y=x$, $xrightarrow 0$ then I get the limit is $$frac{1}{2^{frac{a+b}{2}}}$$
and I let $y=0,xrightarrow 0$,I get the limit equal $0$ ,am I right?
Thanks for your help!
real-analysis calculus functional-analysis limits analysis
asked Jan 8 at 6:02
jacksonjackson
1379
$endgroup$
add a comment |
$begingroup$
Suppose $a>0,b>0,r=sqrt{x^2+y^2}$
1)proof $x^ay^b=o(r^l)(rrightarrow 0),0<l<a+b$
$2)lim_{(x,y)rightarrow (0,0)}frac{x^ay^b}{r^{a+b}} $exist or not
My attempt
1)$$frac{x^ay^b}{r^l}=frac{x^ay^b}{(x^2+y^2)^frac{l}{2}}le frac{1}{2^l}frac{x^ay^b}{(xy)^{frac{l}{2}}}$$
And I don’t know what to do next
2)let $y=x$, $xrightarrow 0$ then I get the limit is $$frac{1}{2^{frac{a+b}{2}}}$$
and I let $y=0,xrightarrow 0$,I get the limit equal $0$ ,am I right?
Thanks for your help!
real-analysis calculus functional-analysis limits analysis
asked Jan 8 at 6:02
jacksonjackson
1379
$endgroup$
Suppose $a>0,b>0,r=sqrt{x^2+y^2}$
1)proof $x^ay^b=o(r^l)(rrightarrow 0),0<l<a+b$
$2)lim_{(x,y)rightarrow (0,0)}frac{x^ay^b}{r^{a+b}} $exist or not
My attempt
1)$$frac{x^ay^b}{r^l}=frac{x^ay^b}{(x^2+y^2)^frac{l}{2}}le frac{1}{2^l}frac{x^ay^b}{(xy)^{frac{l}{2}}}$$
And I don’t know what to do next
2)let $y=x$, $xrightarrow 0$ then I get the limit is $$frac{1}{2^{frac{a+b}{2}}}$$
and I let $y=0,xrightarrow 0$,I get the limit equal $0$ ,am I right?
Thanks for your help!
real-analysis calculus functional-analysis limits analysis
real-analysis calculus functional-analysis limits analysis
asked Jan 8 at 6:02
jacksonjackson
1379
asked Jan 8 at 6:02
jacksonjackson
1379
asked Jan 8 at 6:02
jacksonjackson
1379
asked Jan 8 at 6:02
jacksonjackson
1379
asked Jan 8 at 6:02
jacksonjackson
1379
1379
add a comment |
add a comment |
1 Answer
1
active
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votes
$begingroup$
Your answer for 2) is correct. For i) use polar coordinates. $frac {|x|^{a}|y|^{b}} {r^{ l }}=r^{a+b-l} (|cos (theta)|)^{a} (|sin (theta)|)^{b}) to 0$.
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your answer for 2) is correct. For i) use polar coordinates. $frac {|x|^{a}|y|^{b}} {r^{ l }}=r^{a+b-l} (|cos (theta)|)^{a} (|sin (theta)|)^{b}) to 0$.
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
add a comment |
$begingroup$
Your answer for 2) is correct. For i) use polar coordinates. $frac {|x|^{a}|y|^{b}} {r^{ l }}=r^{a+b-l} (|cos (theta)|)^{a} (|sin (theta)|)^{b}) to 0$.
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
add a comment |
$begingroup$
Your answer for 2) is correct. For i) use polar coordinates. $frac {|x|^{a}|y|^{b}} {r^{ l }}=r^{a+b-l} (|cos (theta)|)^{a} (|sin (theta)|)^{b}) to 0$.
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
Your answer for 2) is correct. For i) use polar coordinates. $frac {|x|^{a}|y|^{b}} {r^{ l }}=r^{a+b-l} (|cos (theta)|)^{a} (|sin (theta)|)^{b}) to 0$.
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
edited Jan 8 at 6:17
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 6:12
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
add a comment |
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
Sin($theta)$’s power should be b?
$endgroup$
– jackson
Jan 8 at 6:15
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
$begingroup$
@jackson Sure! Corrected.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 6:18
add a comment |
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