Show that $ lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists} iff...
$begingroup$
Ted Shifrin on this site posed an interesting problem to me: show that
$$ lim_{(x,y) to (0,0)} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists} iff fracalphagamma + fracbetadelta > 1, ,,,,text{where } alpha, beta, gamma, delta >0$$
I think I've got the $(Leftarrow)$ direction as follows: assume WLOG that $gamma leq delta$. Then switch to polar coordinates and get
$$frac {r^ {alpha + beta} (*) } {r^gamma (1(*) + r^{delta - gamma}(*))},$$
where $(*)$ represents some trig stuff that is bounded near zero.
Edit: I need to make sure that $(1(*) + r^{delta - gamma}(*))$ is bounded below here.
The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
calculus limits multivariable-calculus
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
$endgroup$
add a comment |
$begingroup$
Ted Shifrin on this site posed an interesting problem to me: show that
$$ lim_{(x,y) to (0,0)} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists} iff fracalphagamma + fracbetadelta > 1, ,,,,text{where } alpha, beta, gamma, delta >0$$
I think I've got the $(Leftarrow)$ direction as follows: assume WLOG that $gamma leq delta$. Then switch to polar coordinates and get
$$frac {r^ {alpha + beta} (*) } {r^gamma (1(*) + r^{delta - gamma}(*))},$$
where $(*)$ represents some trig stuff that is bounded near zero.
Edit: I need to make sure that $(1(*) + r^{delta - gamma}(*))$ is bounded below here.
The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
calculus limits multivariable-calculus
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
$endgroup$
$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
$endgroup$
– M.H
Jul 25 '13 at 13:21
$begingroup$
Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
1
$begingroup$
For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
$endgroup$
– Pedro M.
Jul 25 '13 at 14:07
$begingroup$
That's a good point. I need to fix some of the details there.
$endgroup$
– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
$endgroup$
– Pedro M.
Jul 25 '13 at 14:52
add a comment |
$begingroup$
Ted Shifrin on this site posed an interesting problem to me: show that
$$ lim_{(x,y) to (0,0)} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists} iff fracalphagamma + fracbetadelta > 1, ,,,,text{where } alpha, beta, gamma, delta >0$$
I think I've got the $(Leftarrow)$ direction as follows: assume WLOG that $gamma leq delta$. Then switch to polar coordinates and get
$$frac {r^ {alpha + beta} (*) } {r^gamma (1(*) + r^{delta - gamma}(*))},$$
where $(*)$ represents some trig stuff that is bounded near zero.
Edit: I need to make sure that $(1(*) + r^{delta - gamma}(*))$ is bounded below here.
The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
calculus limits multivariable-calculus
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
$endgroup$
Ted Shifrin on this site posed an interesting problem to me: show that
$$ lim_{(x,y) to (0,0)} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists} iff fracalphagamma + fracbetadelta > 1, ,,,,text{where } alpha, beta, gamma, delta >0$$
I think I've got the $(Leftarrow)$ direction as follows: assume WLOG that $gamma leq delta$. Then switch to polar coordinates and get
$$frac {r^ {alpha + beta} (*) } {r^gamma (1(*) + r^{delta - gamma}(*))},$$
where $(*)$ represents some trig stuff that is bounded near zero.
Edit: I need to make sure that $(1(*) + r^{delta - gamma}(*))$ is bounded below here.
The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
calculus limits multivariable-calculus
calculus limits multivariable-calculus
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
edited Jul 25 '13 at 14:18
Eric Auld
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
asked Jul 25 '13 at 13:15
Eric AuldEric Auld
12.8k431109
12.8k431109
$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
$endgroup$
– M.H
Jul 25 '13 at 13:21
$begingroup$
Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
1
$begingroup$
For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
$endgroup$
– Pedro M.
Jul 25 '13 at 14:07
$begingroup$
That's a good point. I need to fix some of the details there.
$endgroup$
– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
$endgroup$
– Pedro M.
Jul 25 '13 at 14:52
add a comment |
$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
$endgroup$
– M.H
Jul 25 '13 at 13:21
$begingroup$
Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
1
$begingroup$
For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
$endgroup$
– Pedro M.
Jul 25 '13 at 14:07
$begingroup$
That's a good point. I need to fix some of the details there.
$endgroup$
– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
$endgroup$
– Pedro M.
Jul 25 '13 at 14:52
$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
$endgroup$
– M.H
Jul 25 '13 at 13:21
$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
$endgroup$
– M.H
Jul 25 '13 at 13:21
$begingroup$
Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
$begingroup$
Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
1
1
$begingroup$
For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
$endgroup$
– Pedro M.
Jul 25 '13 at 14:07
$begingroup$
For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
$endgroup$
– Pedro M.
Jul 25 '13 at 14:07
$begingroup$
That's a good point. I need to fix some of the details there.
$endgroup$
– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
That's a good point. I need to fix some of the details there.
$endgroup$
– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
$endgroup$
– Pedro M.
Jul 25 '13 at 14:52
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
$endgroup$
– Pedro M.
Jul 25 '13 at 14:52
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
For the $(Leftarrow)$ direction, we first note that
$$lim_{(x,y) to 0} frac{|x|^alpha |y|^beta}{|x|^gamma + |y|^delta} = lim_{r_1,r_2 to 0^+} frac{r^alpha_1 r^beta_2}{r^gamma_1 + r^delta_2} = lim_{t_1,t_2 to 0^+} frac{t^{alphadelta}_1 t^{betagamma}_2}{t^{gammadelta}_1 + t^{gammadelta}_2}, $$
since a sequence $(t_n, s_n) to (0,0)$ if and only if $(t^delta_n, s^gamma_n) to (0,0)$
Now for the rightmost limit, your basic argument works: by switching to polar coordinates we have
$$lim_{r to 0} frac{r^{alpha delta + beta gamma}left|sin(theta)^{(beta gamma)} cos(theta)^{(alpha delta)}right|}{r^{gamma delta} left(left|sin(theta)^{(gamma delta)}right| + left|cos(theta)^{(gamma delta)}right|right)}, $$
which we can easily show to converge (to zero) if $alpha delta + beta gamma > gamma delta$.
For the ($Rightarrow$) direction, assume that the said limit exists. Then, in particular, this limit also exists:
$$lim_{t to 0^+} frac{(t^delta)^alpha (t^gamma)^delta}{(t^delta)^gamma + (t^gamma)^delta},$$
which clearly exists only if $alpha delta + beta gamma > gamma delta$.
EDIT: The last limit exists if $alpha delta + beta gamma geq gamma delta$.
To show that we can't have $alpha delta + beta gamma = gamma delta$, notice that while in this case the above limit is $frac12$, we also have other "subsequences" converging to zero: let $N$ be such that $beta N > gamma$ and $delta N > gamma$. Then
$$ frac{t^alpha (t^N)^beta}{t^gamma + (t^N)^delta} approx frac{t^alpha (t^N)^beta}{t^gamma} to 0 $$
Thus the limit does not exist when $alpha delta + beta gamma = gamma delta.$
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
$endgroup$
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
|
show 2 more comments
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$begingroup$
For the $(Leftarrow)$ direction, we first note that
$$lim_{(x,y) to 0} frac{|x|^alpha |y|^beta}{|x|^gamma + |y|^delta} = lim_{r_1,r_2 to 0^+} frac{r^alpha_1 r^beta_2}{r^gamma_1 + r^delta_2} = lim_{t_1,t_2 to 0^+} frac{t^{alphadelta}_1 t^{betagamma}_2}{t^{gammadelta}_1 + t^{gammadelta}_2}, $$
since a sequence $(t_n, s_n) to (0,0)$ if and only if $(t^delta_n, s^gamma_n) to (0,0)$
Now for the rightmost limit, your basic argument works: by switching to polar coordinates we have
$$lim_{r to 0} frac{r^{alpha delta + beta gamma}left|sin(theta)^{(beta gamma)} cos(theta)^{(alpha delta)}right|}{r^{gamma delta} left(left|sin(theta)^{(gamma delta)}right| + left|cos(theta)^{(gamma delta)}right|right)}, $$
which we can easily show to converge (to zero) if $alpha delta + beta gamma > gamma delta$.
For the ($Rightarrow$) direction, assume that the said limit exists. Then, in particular, this limit also exists:
$$lim_{t to 0^+} frac{(t^delta)^alpha (t^gamma)^delta}{(t^delta)^gamma + (t^gamma)^delta},$$
which clearly exists only if $alpha delta + beta gamma > gamma delta$.
EDIT: The last limit exists if $alpha delta + beta gamma geq gamma delta$.
To show that we can't have $alpha delta + beta gamma = gamma delta$, notice that while in this case the above limit is $frac12$, we also have other "subsequences" converging to zero: let $N$ be such that $beta N > gamma$ and $delta N > gamma$. Then
$$ frac{t^alpha (t^N)^beta}{t^gamma + (t^N)^delta} approx frac{t^alpha (t^N)^beta}{t^gamma} to 0 $$
Thus the limit does not exist when $alpha delta + beta gamma = gamma delta.$
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
$endgroup$
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
|
show 2 more comments
$begingroup$
For the $(Leftarrow)$ direction, we first note that
$$lim_{(x,y) to 0} frac{|x|^alpha |y|^beta}{|x|^gamma + |y|^delta} = lim_{r_1,r_2 to 0^+} frac{r^alpha_1 r^beta_2}{r^gamma_1 + r^delta_2} = lim_{t_1,t_2 to 0^+} frac{t^{alphadelta}_1 t^{betagamma}_2}{t^{gammadelta}_1 + t^{gammadelta}_2}, $$
since a sequence $(t_n, s_n) to (0,0)$ if and only if $(t^delta_n, s^gamma_n) to (0,0)$
Now for the rightmost limit, your basic argument works: by switching to polar coordinates we have
$$lim_{r to 0} frac{r^{alpha delta + beta gamma}left|sin(theta)^{(beta gamma)} cos(theta)^{(alpha delta)}right|}{r^{gamma delta} left(left|sin(theta)^{(gamma delta)}right| + left|cos(theta)^{(gamma delta)}right|right)}, $$
which we can easily show to converge (to zero) if $alpha delta + beta gamma > gamma delta$.
For the ($Rightarrow$) direction, assume that the said limit exists. Then, in particular, this limit also exists:
$$lim_{t to 0^+} frac{(t^delta)^alpha (t^gamma)^delta}{(t^delta)^gamma + (t^gamma)^delta},$$
which clearly exists only if $alpha delta + beta gamma > gamma delta$.
EDIT: The last limit exists if $alpha delta + beta gamma geq gamma delta$.
To show that we can't have $alpha delta + beta gamma = gamma delta$, notice that while in this case the above limit is $frac12$, we also have other "subsequences" converging to zero: let $N$ be such that $beta N > gamma$ and $delta N > gamma$. Then
$$ frac{t^alpha (t^N)^beta}{t^gamma + (t^N)^delta} approx frac{t^alpha (t^N)^beta}{t^gamma} to 0 $$
Thus the limit does not exist when $alpha delta + beta gamma = gamma delta.$
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
$endgroup$
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
|
show 2 more comments
$begingroup$
For the $(Leftarrow)$ direction, we first note that
$$lim_{(x,y) to 0} frac{|x|^alpha |y|^beta}{|x|^gamma + |y|^delta} = lim_{r_1,r_2 to 0^+} frac{r^alpha_1 r^beta_2}{r^gamma_1 + r^delta_2} = lim_{t_1,t_2 to 0^+} frac{t^{alphadelta}_1 t^{betagamma}_2}{t^{gammadelta}_1 + t^{gammadelta}_2}, $$
since a sequence $(t_n, s_n) to (0,0)$ if and only if $(t^delta_n, s^gamma_n) to (0,0)$
Now for the rightmost limit, your basic argument works: by switching to polar coordinates we have
$$lim_{r to 0} frac{r^{alpha delta + beta gamma}left|sin(theta)^{(beta gamma)} cos(theta)^{(alpha delta)}right|}{r^{gamma delta} left(left|sin(theta)^{(gamma delta)}right| + left|cos(theta)^{(gamma delta)}right|right)}, $$
which we can easily show to converge (to zero) if $alpha delta + beta gamma > gamma delta$.
For the ($Rightarrow$) direction, assume that the said limit exists. Then, in particular, this limit also exists:
$$lim_{t to 0^+} frac{(t^delta)^alpha (t^gamma)^delta}{(t^delta)^gamma + (t^gamma)^delta},$$
which clearly exists only if $alpha delta + beta gamma > gamma delta$.
EDIT: The last limit exists if $alpha delta + beta gamma geq gamma delta$.
To show that we can't have $alpha delta + beta gamma = gamma delta$, notice that while in this case the above limit is $frac12$, we also have other "subsequences" converging to zero: let $N$ be such that $beta N > gamma$ and $delta N > gamma$. Then
$$ frac{t^alpha (t^N)^beta}{t^gamma + (t^N)^delta} approx frac{t^alpha (t^N)^beta}{t^gamma} to 0 $$
Thus the limit does not exist when $alpha delta + beta gamma = gamma delta.$
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
$endgroup$
For the $(Leftarrow)$ direction, we first note that
$$lim_{(x,y) to 0} frac{|x|^alpha |y|^beta}{|x|^gamma + |y|^delta} = lim_{r_1,r_2 to 0^+} frac{r^alpha_1 r^beta_2}{r^gamma_1 + r^delta_2} = lim_{t_1,t_2 to 0^+} frac{t^{alphadelta}_1 t^{betagamma}_2}{t^{gammadelta}_1 + t^{gammadelta}_2}, $$
since a sequence $(t_n, s_n) to (0,0)$ if and only if $(t^delta_n, s^gamma_n) to (0,0)$
Now for the rightmost limit, your basic argument works: by switching to polar coordinates we have
$$lim_{r to 0} frac{r^{alpha delta + beta gamma}left|sin(theta)^{(beta gamma)} cos(theta)^{(alpha delta)}right|}{r^{gamma delta} left(left|sin(theta)^{(gamma delta)}right| + left|cos(theta)^{(gamma delta)}right|right)}, $$
which we can easily show to converge (to zero) if $alpha delta + beta gamma > gamma delta$.
For the ($Rightarrow$) direction, assume that the said limit exists. Then, in particular, this limit also exists:
$$lim_{t to 0^+} frac{(t^delta)^alpha (t^gamma)^delta}{(t^delta)^gamma + (t^gamma)^delta},$$
which clearly exists only if $alpha delta + beta gamma > gamma delta$.
EDIT: The last limit exists if $alpha delta + beta gamma geq gamma delta$.
To show that we can't have $alpha delta + beta gamma = gamma delta$, notice that while in this case the above limit is $frac12$, we also have other "subsequences" converging to zero: let $N$ be such that $beta N > gamma$ and $delta N > gamma$. Then
$$ frac{t^alpha (t^N)^beta}{t^gamma + (t^N)^delta} approx frac{t^alpha (t^N)^beta}{t^gamma} to 0 $$
Thus the limit does not exist when $alpha delta + beta gamma = gamma delta.$
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
edited Jul 25 '13 at 17:55
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
answered Jul 25 '13 at 15:38
Pedro M.Pedro M.
2,7631216
2,7631216
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Thanks for your solution! You have no idea how much that problem was bothering me.
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– Eric Auld
Jul 25 '13 at 15:41
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I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
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– Eric Auld
Jul 25 '13 at 17:23
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By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
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– Pedro M.
Jul 25 '13 at 18:01
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Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
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– Eric Auld
Jul 25 '13 at 18:25
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I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
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– Eric Auld
Jul 25 '13 at 18:37
|
show 2 more comments
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
Thanks for your solution! You have no idea how much that problem was bothering me.
$endgroup$
– Eric Auld
Jul 25 '13 at 15:41
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
I notice we can generalize this to $n$ dimensions by noting that the factor in the denominator which is here represented by $|sin(theta)|^{(gamma delta)} + |cos(theta)|^{(gamma delta)}$ (I think we want the bars on the inside, right?) is (in any dimension $mathbb{R}^n$) the $infty$ norm of a vector whose $2/(gammadelta)$-norm is $1$, so it is bounded below for all $theta$.
$endgroup$
– Eric Auld
Jul 25 '13 at 17:23
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
By generalizing to $n$ dimensions do you mean the following? $$lim_{x to 0} frac{x_1^{alpha_1} ldots x_n^{alpha_n}}{x_1^{beta_1} + ldots + x_n^{beta_n}} text{ exists } Leftrightarrow sum_i frac{alpha_i}{beta_i} > 1$$
$endgroup$
– Pedro M.
Jul 25 '13 at 18:01
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
Yes, exactly. I think we'd make the substitution $x_i^{alpha_i} leftrightarrow x_i^{alpha_i prod_{jneq i}beta_j}$.
$endgroup$
– Eric Auld
Jul 25 '13 at 18:25
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
$begingroup$
I note that a consequence of this generalization would be that $$frac { x_1^{alpha_1} dotsb x_n^{alpha_n} } {|x|^p}$$ converges iff $sum alpha_i > p$, since $$ frac1n (|x_1|^p + dotsb + |x_n|^p) leq |x|^p leq n^{p/2}(|x_1|^p + dotsb + |x_n|^p). $$
$endgroup$
– Eric Auld
Jul 25 '13 at 18:37
|
show 2 more comments
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$begingroup$
Hint:if we let $large{lim_{(x,y) to 0} frac {|x|^ alpha |y|^ beta} {|x|^ gamma + |y|^ delta} text {exists}}$ exactly its limit in all path is exist know let $x=y $
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– M.H
Jul 25 '13 at 13:21
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Right, if we let $x=2y$, say, then what do we do with the denominator?
$endgroup$
– Eric Auld
Jul 25 '13 at 13:33
1
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For the $Leftarrow$ direction, I think you need to elaborate on "some trig stuff that is bounded near zero". For the numerator this is enough, but you need to prove that the denominator is bounded below or something of the sort. In fact, I think you cannot prove this by ignoring the trigonometric parts: otherwise, you would have proven that $$frac{alpha + beta}{min(gamma, delta)} > 1$$ sufficed for the limit to exist.
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– Pedro M.
Jul 25 '13 at 14:07
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That's a good point. I need to fix some of the details there.
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– Eric Auld
Jul 25 '13 at 14:15
$begingroup$
By the way, the $Rightarrow$ direction is actually easier. Suppose the said limit exists, and consider the particular case where $|x| = t^delta, |y| = t^gamma$ ($t to 0$). What do you get?
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– Pedro M.
Jul 25 '13 at 14:52