Mixing
Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.
$begingroup$
Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.
By the answers only the latter has a limit at $(0,0)$ but I don't know how to prove.
real-analysis limits multivariable-calculus exponentiation
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
$endgroup$
|
show 1 more comment
$begingroup$
Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.
By the answers only the latter has a limit at $(0,0)$ but I don't know how to prove.
real-analysis limits multivariable-calculus exponentiation
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
$endgroup$
$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:50
$begingroup$
@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
$endgroup$
– helios321
Jan 18 at 5:12
$begingroup$
The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:14
$begingroup$
@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
$endgroup$
– helios321
Jan 18 at 5:24
$begingroup$
$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:26
|
show 1 more comment
$begingroup$
Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.
By the answers only the latter has a limit at $(0,0)$ but I don't know how to prove.
real-analysis limits multivariable-calculus exponentiation
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
$endgroup$
Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.
By the answers only the latter has a limit at $(0,0)$ but I don't know how to prove.
real-analysis limits multivariable-calculus exponentiation
real-analysis limits multivariable-calculus exponentiation
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
edited Jan 18 at 11:12
Martin Sleziak
44.7k10119272
44.7k10119272
asked Jan 18 at 4:06
helios321helios321
56349
asked Jan 18 at 4:06
helios321helios321
56349
asked Jan 18 at 4:06
helios321helios321
56349
56349
$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:50
$begingroup$
@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
$endgroup$
– helios321
Jan 18 at 5:12
$begingroup$
The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:14
$begingroup$
@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
$endgroup$
– helios321
Jan 18 at 5:24
$begingroup$
$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:26
|
show 1 more comment
$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:50
$begingroup$
@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
$endgroup$
– helios321
Jan 18 at 5:12
$begingroup$
The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:14
$begingroup$
@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
$endgroup$
– helios321
Jan 18 at 5:24
$begingroup$
$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:26
$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:50
$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 4:50
$begingroup$
@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
$endgroup$
– helios321
Jan 18 at 5:12
$begingroup$
@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
$endgroup$
– helios321
Jan 18 at 5:12
$begingroup$
The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:14
$begingroup$
The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:14
$begingroup$
@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
$endgroup$
– helios321
Jan 18 at 5:24
$begingroup$
@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
$endgroup$
– helios321
Jan 18 at 5:24
$begingroup$
$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:26
$begingroup$
$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 5:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I might as well write an answer.
For the first question, we note that if $lim_{(x,y) to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n to 0,y_n to 0$ we should have $f(x_n,y_n) to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,frac 1n)$ we have $f(x_n,y_n) = 0^{frac 1n} = 0 to 0$ ,and with $(x_n,y_n) = (frac 1n,0)$ we have $f(x_n,y_n) = left(frac{1}{n}right)^0 = 1 to 1$, and $0 neq 1$ so no candidate for $L$ can exist.
For the second question, we have tho show that the limit of $|x|^{frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.
To show that the limit is zero, fix $epsilon > 0$. Suppose that $|x| < min{frac 12,epsilon}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $frac 1{|y|} > 1$, which means that $|x|^{frac 1{|y|}} < |x| < epsilon$.
We have shown therefore, that $|x| < min{frac 12,epsilon}$ and $|y|<1$ implies $|x|^{frac 1{|y|}}<epsilon$. Can you do it from here?
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
$endgroup$
– helios321
Jan 18 at 7:55
$begingroup$
Can you produce it with the info given
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
I might as well write an answer.
For the first question, we note that if $lim_{(x,y) to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n to 0,y_n to 0$ we should have $f(x_n,y_n) to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,frac 1n)$ we have $f(x_n,y_n) = 0^{frac 1n} = 0 to 0$ ,and with $(x_n,y_n) = (frac 1n,0)$ we have $f(x_n,y_n) = left(frac{1}{n}right)^0 = 1 to 1$, and $0 neq 1$ so no candidate for $L$ can exist.
For the second question, we have tho show that the limit of $|x|^{frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.
To show that the limit is zero, fix $epsilon > 0$. Suppose that $|x| < min{frac 12,epsilon}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $frac 1{|y|} > 1$, which means that $|x|^{frac 1{|y|}} < |x| < epsilon$.
We have shown therefore, that $|x| < min{frac 12,epsilon}$ and $|y|<1$ implies $|x|^{frac 1{|y|}}<epsilon$. Can you do it from here?
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
$endgroup$
– helios321
Jan 18 at 7:55
$begingroup$
Can you produce it with the info given
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
|
show 1 more comment
$begingroup$
I might as well write an answer.
For the first question, we note that if $lim_{(x,y) to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n to 0,y_n to 0$ we should have $f(x_n,y_n) to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,frac 1n)$ we have $f(x_n,y_n) = 0^{frac 1n} = 0 to 0$ ,and with $(x_n,y_n) = (frac 1n,0)$ we have $f(x_n,y_n) = left(frac{1}{n}right)^0 = 1 to 1$, and $0 neq 1$ so no candidate for $L$ can exist.
For the second question, we have tho show that the limit of $|x|^{frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.
To show that the limit is zero, fix $epsilon > 0$. Suppose that $|x| < min{frac 12,epsilon}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $frac 1{|y|} > 1$, which means that $|x|^{frac 1{|y|}} < |x| < epsilon$.
We have shown therefore, that $|x| < min{frac 12,epsilon}$ and $|y|<1$ implies $|x|^{frac 1{|y|}}<epsilon$. Can you do it from here?
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
$endgroup$
– helios321
Jan 18 at 7:55
$begingroup$
Can you produce it with the info given
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
|
show 1 more comment
$begingroup$
I might as well write an answer.
For the first question, we note that if $lim_{(x,y) to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n to 0,y_n to 0$ we should have $f(x_n,y_n) to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,frac 1n)$ we have $f(x_n,y_n) = 0^{frac 1n} = 0 to 0$ ,and with $(x_n,y_n) = (frac 1n,0)$ we have $f(x_n,y_n) = left(frac{1}{n}right)^0 = 1 to 1$, and $0 neq 1$ so no candidate for $L$ can exist.
For the second question, we have tho show that the limit of $|x|^{frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.
To show that the limit is zero, fix $epsilon > 0$. Suppose that $|x| < min{frac 12,epsilon}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $frac 1{|y|} > 1$, which means that $|x|^{frac 1{|y|}} < |x| < epsilon$.
We have shown therefore, that $|x| < min{frac 12,epsilon}$ and $|y|<1$ implies $|x|^{frac 1{|y|}}<epsilon$. Can you do it from here?
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
I might as well write an answer.
For the first question, we note that if $lim_{(x,y) to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n to 0,y_n to 0$ we should have $f(x_n,y_n) to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,frac 1n)$ we have $f(x_n,y_n) = 0^{frac 1n} = 0 to 0$ ,and with $(x_n,y_n) = (frac 1n,0)$ we have $f(x_n,y_n) = left(frac{1}{n}right)^0 = 1 to 1$, and $0 neq 1$ so no candidate for $L$ can exist.
For the second question, we have tho show that the limit of $|x|^{frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.
To show that the limit is zero, fix $epsilon > 0$. Suppose that $|x| < min{frac 12,epsilon}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $frac 1{|y|} > 1$, which means that $|x|^{frac 1{|y|}} < |x| < epsilon$.
We have shown therefore, that $|x| < min{frac 12,epsilon}$ and $|y|<1$ implies $|x|^{frac 1{|y|}}<epsilon$. Can you do it from here?
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
answered Jan 18 at 7:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
38.8k33477
$begingroup$
Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
$endgroup$
– helios321
Jan 18 at 7:55
$begingroup$
Can you produce it with the info given
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
|
show 1 more comment
$begingroup$
Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
$endgroup$
– helios321
Jan 18 at 7:55
$begingroup$
Can you produce it with the info given
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
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Exactly, use that relation
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– астон вілла олоф мэллбэрг
Jan 18 at 11:45
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Have you managed to complete the argument?
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– астон вілла олоф мэллбэрг
Jan 19 at 6:47
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Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
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– helios321
Jan 18 at 7:55
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Don't you need to find $delta$ such that $sqrt{x^2+y^2}<delta$ gives you that inequality?
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– helios321
Jan 18 at 7:55
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Can you produce it with the info given
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– астон вілла олоф мэллбэрг
Jan 18 at 9:45
$begingroup$
Can you produce it with the info given
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– астон вілла олоф мэллбэрг
Jan 18 at 9:45
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Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
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– helios321
Jan 18 at 10:01
$begingroup$
Do you mean to use the fact that $sqrt{x^2+y^2} leq |x| + |y|$?
$endgroup$
– helios321
Jan 18 at 10:01
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Exactly, use that relation
$endgroup$
– астон вілла олоф мэллбэрг
Jan 18 at 11:45
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
$begingroup$
Have you managed to complete the argument?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 6:47
|
show 1 more comment
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$begingroup$
Have you seen examples of "how to show that a function does not have a limit at a point" with any function at any point?
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– астон вілла олоф мэллбэрг
Jan 18 at 4:50
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@астонвіллаолофмэллбэрг Yes but I can't seem to deal with these absolute value functions.
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– helios321
Jan 18 at 5:12
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The key point in the proofs involving non-existence of a limit, is to show that along some path to the point, either the limit does not exist, or showing that the limit along two different paths is a different value. For example, for the first question, if you approach $(0,0)$ via the path $x = 0$ (or via $x$-axis), then $f(x,y)$ is always zero. On the other hand, if we approach via the y-axis,...
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– астон вілла олоф мэллбэрг
Jan 18 at 5:14
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@астонвіллаолофмэллбэрг it must be because im not sure how to deal with $0^x$ and $x^0$ as limits as $x$ approaches $0$ because im confusing it with $x^x$ which has limit $1$. Can you clarify this.
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– helios321
Jan 18 at 5:24
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$x^0 = 1$ for all $x neq 0$, and $0^x =0$ for all $x neq 0$. You don't need to bring in $x^x$ into this argument : along the $x$ axis, the limit is $lim_{y to 0} 0^{|y|} = 0$, and along the $y$ axis the limit is $lim_{x to 0} |x|^0 = 1$, and since $0 neq 1$ the limit does not exist
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– астон вілла олоф мэллбэрг
Jan 18 at 5:26