Delta in Epsilon delta definition of limit [closed] The Next CEO of Stack OverflowLimits using...
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Delta in Epsilon delta definition of limit [closed]
The Next CEO of Stack OverflowLimits using epsilon delta definition $f(x,y)=xy$ for functions of two variables$epsilon - delta$ definition of a limitProving a limit using epsilon delta definitionInterpretation of $epsilon$-$delta$ limit definitiondoes epsilon-delta definition of limit presuppose that the function is defined everywhere at (x-delta,x+delta)?Question about limits using delta - epsilon definition$epsilon-delta$ definition for limits involving $infty$($epsilon$,$delta$) Definition of limit$epsilon−delta$ limit proofLimit with Epsilon - Delta method
$begingroup$
We know delta in epsilon delta definition of limit depends upon epsilon.
Can delta in epsilon delta definition of limits depend upon x also?
limits
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
$endgroup$
closed as unclear what you're asking by Thomas Shelby, Alex Provost, Cesareo, mrtaurho, José Carlos Santos Mar 17 at 13:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
We know delta in epsilon delta definition of limit depends upon epsilon.
Can delta in epsilon delta definition of limits depend upon x also?
limits
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
$endgroup$
closed as unclear what you're asking by Thomas Shelby, Alex Provost, Cesareo, mrtaurho, José Carlos Santos Mar 17 at 13:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
1
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21
add a comment |
$begingroup$
We know delta in epsilon delta definition of limit depends upon epsilon.
Can delta in epsilon delta definition of limits depend upon x also?
limits
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
$endgroup$
We know delta in epsilon delta definition of limit depends upon epsilon.
Can delta in epsilon delta definition of limits depend upon x also?
limits
limits
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
asked Mar 16 at 17:16
Sukh JotSukh Jot
11
11
closed as unclear what you're asking by Thomas Shelby, Alex Provost, Cesareo, mrtaurho, José Carlos Santos Mar 17 at 13:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Thomas Shelby, Alex Provost, Cesareo, mrtaurho, José Carlos Santos Mar 17 at 13:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
1
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21
add a comment |
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
1
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
1
1
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, and it often does!
For example, let's say that $f$ is a real-valued function on some subset $E$ of the real line. We say that $f$ is continuous if for every $epsilon>0$ and every $xin E,$ there is some $delta>0$ such that, whenever $yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ This $delta,$ though, may be different for different $x,$ even when $epsilon$ is the same. For instance, consider $f(x)=frac1x$ and $E={xinBbb R:xne 0}.$ We can show that $f$ is continuous, but we'll need to choose smaller $delta$-values as $x$ gets closer to $0,$ since $f$ is changing so rapidly there.
Now, we say that $f$ is uniformly continuous if for every $epsilon>0,$ there is some $delta>0$ such that, whenever $x,yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ In this case, the $delta$ does not depend on $x.$ Bringing it back to the function $f(x)=frac1x,$ but with $E=[1,infty),$ instead, it turns out that $f$ is uniformly continuous on $E$! Since $f$ changes less and less quickly the farther $x$ gets from $0,$ all we need to do is find a $delta$ that works when $x=1,$ and it will work for all other $xin E,$ as well!
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, and it often does!
For example, let's say that $f$ is a real-valued function on some subset $E$ of the real line. We say that $f$ is continuous if for every $epsilon>0$ and every $xin E,$ there is some $delta>0$ such that, whenever $yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ This $delta,$ though, may be different for different $x,$ even when $epsilon$ is the same. For instance, consider $f(x)=frac1x$ and $E={xinBbb R:xne 0}.$ We can show that $f$ is continuous, but we'll need to choose smaller $delta$-values as $x$ gets closer to $0,$ since $f$ is changing so rapidly there.
Now, we say that $f$ is uniformly continuous if for every $epsilon>0,$ there is some $delta>0$ such that, whenever $x,yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ In this case, the $delta$ does not depend on $x.$ Bringing it back to the function $f(x)=frac1x,$ but with $E=[1,infty),$ instead, it turns out that $f$ is uniformly continuous on $E$! Since $f$ changes less and less quickly the farther $x$ gets from $0,$ all we need to do is find a $delta$ that works when $x=1,$ and it will work for all other $xin E,$ as well!
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
$endgroup$
add a comment |
$begingroup$
Yes, and it often does!
For example, let's say that $f$ is a real-valued function on some subset $E$ of the real line. We say that $f$ is continuous if for every $epsilon>0$ and every $xin E,$ there is some $delta>0$ such that, whenever $yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ This $delta,$ though, may be different for different $x,$ even when $epsilon$ is the same. For instance, consider $f(x)=frac1x$ and $E={xinBbb R:xne 0}.$ We can show that $f$ is continuous, but we'll need to choose smaller $delta$-values as $x$ gets closer to $0,$ since $f$ is changing so rapidly there.
Now, we say that $f$ is uniformly continuous if for every $epsilon>0,$ there is some $delta>0$ such that, whenever $x,yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ In this case, the $delta$ does not depend on $x.$ Bringing it back to the function $f(x)=frac1x,$ but with $E=[1,infty),$ instead, it turns out that $f$ is uniformly continuous on $E$! Since $f$ changes less and less quickly the farther $x$ gets from $0,$ all we need to do is find a $delta$ that works when $x=1,$ and it will work for all other $xin E,$ as well!
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
$endgroup$
add a comment |
$begingroup$
Yes, and it often does!
For example, let's say that $f$ is a real-valued function on some subset $E$ of the real line. We say that $f$ is continuous if for every $epsilon>0$ and every $xin E,$ there is some $delta>0$ such that, whenever $yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ This $delta,$ though, may be different for different $x,$ even when $epsilon$ is the same. For instance, consider $f(x)=frac1x$ and $E={xinBbb R:xne 0}.$ We can show that $f$ is continuous, but we'll need to choose smaller $delta$-values as $x$ gets closer to $0,$ since $f$ is changing so rapidly there.
Now, we say that $f$ is uniformly continuous if for every $epsilon>0,$ there is some $delta>0$ such that, whenever $x,yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ In this case, the $delta$ does not depend on $x.$ Bringing it back to the function $f(x)=frac1x,$ but with $E=[1,infty),$ instead, it turns out that $f$ is uniformly continuous on $E$! Since $f$ changes less and less quickly the farther $x$ gets from $0,$ all we need to do is find a $delta$ that works when $x=1,$ and it will work for all other $xin E,$ as well!
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
$endgroup$
Yes, and it often does!
For example, let's say that $f$ is a real-valued function on some subset $E$ of the real line. We say that $f$ is continuous if for every $epsilon>0$ and every $xin E,$ there is some $delta>0$ such that, whenever $yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ This $delta,$ though, may be different for different $x,$ even when $epsilon$ is the same. For instance, consider $f(x)=frac1x$ and $E={xinBbb R:xne 0}.$ We can show that $f$ is continuous, but we'll need to choose smaller $delta$-values as $x$ gets closer to $0,$ since $f$ is changing so rapidly there.
Now, we say that $f$ is uniformly continuous if for every $epsilon>0,$ there is some $delta>0$ such that, whenever $x,yin E$ and $|x-y|<delta,$ we have $bigl|f(x)-f(y)bigr|<epsilon.$ In this case, the $delta$ does not depend on $x.$ Bringing it back to the function $f(x)=frac1x,$ but with $E=[1,infty),$ instead, it turns out that $f$ is uniformly continuous on $E$! Since $f$ changes less and less quickly the farther $x$ gets from $0,$ all we need to do is find a $delta$ that works when $x=1,$ and it will work for all other $xin E,$ as well!
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
edited Mar 16 at 18:01
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
answered Mar 16 at 17:24
Cameron BuieCameron Buie
86.3k773161
86.3k773161
add a comment |
add a comment |
$begingroup$
The value of $delta$ depends on the value of $x$ and $epsilon.$
$endgroup$
– mfl
Mar 16 at 17:19
$begingroup$
Yes, it depends on $x$ in general.
$endgroup$
– GReyes
Mar 16 at 17:19
1
$begingroup$
When it does not depend on $x$, it leads to the notion of uniform continuity, convergence, &c.
$endgroup$
– Bernard
Mar 16 at 17:21
$begingroup$
When it does not depend on $x$, we say that the convergence is uniform. This is an important property. [Crossed with Bernard]
$endgroup$
– Yves Daoust
Mar 16 at 17:21