Can I prove that a 2-variable limit does not exists if the limit on a curve is infinity?
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Consider a 2 variable function $f(x,y)$ and the limit
$$lim_{(x,y)to (0,0)} f(x,y)$$
If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and
$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$
Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?
I'm not sure about this because of the infinity in one of the two limits above
calculus limits multivariable-calculus infinity
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
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add a comment |
$begingroup$
Consider a 2 variable function $f(x,y)$ and the limit
$$lim_{(x,y)to (0,0)} f(x,y)$$
If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and
$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$
Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?
I'm not sure about this because of the infinity in one of the two limits above
calculus limits multivariable-calculus infinity
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
$endgroup$
2
$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36
add a comment |
$begingroup$
Consider a 2 variable function $f(x,y)$ and the limit
$$lim_{(x,y)to (0,0)} f(x,y)$$
If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and
$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$
Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?
I'm not sure about this because of the infinity in one of the two limits above
calculus limits multivariable-calculus infinity
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
$endgroup$
Consider a 2 variable function $f(x,y)$ and the limit
$$lim_{(x,y)to (0,0)} f(x,y)$$
If I find two continuous functions $gamma_1(t)$ and $gamma_2(t)$ such that $gamma_1(0)=gamma_2(0)=(0,0)$ and
$$lim_{tto 0} f(gamma_1(t))=l in mathbb{R} ,,,,,,,,,,, and ,,,,,,,,,, lim_{tto 0} f(gamma_2(t))=+infty $$
Can I conclude that the original limit $lim_{(x,y)to (0,0)} f(x,y)$ does not exist?
I'm not sure about this because of the infinity in one of the two limits above
calculus limits multivariable-calculus infinity
calculus limits multivariable-calculus infinity
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
asked Dec 6 '18 at 22:34
GianolepoGianolepo
790919
790919
2
$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36
add a comment |
2
$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36
2
2
$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36
$begingroup$
yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36
add a comment |
1 Answer
1
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Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for
$$lim_{(x,y)to (0,0)} frac{xy}{x^3-y^3} $$
we have
$frac{xy}{x^3-y^3} to 0$ for $x=0$
$frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
$endgroup$
add a comment |
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$begingroup$
Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for
$$lim_{(x,y)to (0,0)} frac{xy}{x^3-y^3} $$
we have
$frac{xy}{x^3-y^3} to 0$ for $x=0$
$frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for
$$lim_{(x,y)to (0,0)} frac{xy}{x^3-y^3} $$
we have
$frac{xy}{x^3-y^3} to 0$ for $x=0$
$frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for
$$lim_{(x,y)to (0,0)} frac{xy}{x^3-y^3} $$
we have
$frac{xy}{x^3-y^3} to 0$ for $x=0$
$frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
$endgroup$
Yes of course it suffices to find, at least, two different limits finite or infinite, for different paths as for example for
$$lim_{(x,y)to (0,0)} frac{xy}{x^3-y^3} $$
we have
$frac{xy}{x^3-y^3} to 0$ for $x=0$
$frac{xy}{x^3-y^3} to +infty$ for $y=-x=t to 0^+$
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
edited Dec 6 '18 at 23:04
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
answered Dec 6 '18 at 22:53
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
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yes of course it doesn't exist !
$endgroup$
– Surb
Dec 6 '18 at 22:36