Epsilon-Delta Proof of Limits Being Equal
up vote
0
down vote
favorite
How would I go about proving that two limits are equal to each other using the Epsilon-Delta definition?
Moreover how can I prove that: $$lim_{xto0}f(x) = lim_{xto a}f(x-a)$$ using the Epsilon-Delta definition? The intuition for this seem clear. However, I have do not know how a formal proof can be developed.
calculus real-analysis limits
asked Oct 9 '17 at 2:53
Michael Connor
399212
add a comment |
up vote
0
down vote
favorite
How would I go about proving that two limits are equal to each other using the Epsilon-Delta definition?
Moreover how can I prove that: $$lim_{xto0}f(x) = lim_{xto a}f(x-a)$$ using the Epsilon-Delta definition? The intuition for this seem clear. However, I have do not know how a formal proof can be developed.
calculus real-analysis limits
asked Oct 9 '17 at 2:53
Michael Connor
399212
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How would I go about proving that two limits are equal to each other using the Epsilon-Delta definition?
Moreover how can I prove that: $$lim_{xto0}f(x) = lim_{xto a}f(x-a)$$ using the Epsilon-Delta definition? The intuition for this seem clear. However, I have do not know how a formal proof can be developed.
calculus real-analysis limits
asked Oct 9 '17 at 2:53
Michael Connor
399212
How would I go about proving that two limits are equal to each other using the Epsilon-Delta definition?
Moreover how can I prove that: $$lim_{xto0}f(x) = lim_{xto a}f(x-a)$$ using the Epsilon-Delta definition? The intuition for this seem clear. However, I have do not know how a formal proof can be developed.
calculus real-analysis limits
calculus real-analysis limits
asked Oct 9 '17 at 2:53
Michael Connor
399212
asked Oct 9 '17 at 2:53
Michael Connor
399212
asked Oct 9 '17 at 2:53
Michael Connor
399212
asked Oct 9 '17 at 2:53
Michael Connor
399212
asked Oct 9 '17 at 2:53
Michael Connor
399212
399212
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31
add a comment |
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
To show at least the second problem, assume that $lim_{xto 0} f(x) = L$. We will prove that $lim_{x to a}f(x-a) = L$.
By definition, we are to show : for all $epsilon > 0$, there exists $delta > 0$ such that
begin{equation}
|y - a |< delta implies |f(y-a) - L| < epsilon tag{1}
end{equation}
Fix one such $epsilon$, say $epsilon_0$.
We know that $lim_{x to 0} f(x) = L$. This means, that for the above $epsilon_0$, there exists some $delta_0 > 0$ such that begin{equation}|x - 0|( = |x|) < delta_0 implies |f(x) - L| < epsilon_0 tag{2}end{equation}
By taking $delta = delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
begin{equation}
|y-a| < delta_0
implies |f(y-a) - L |< epsilon_0 end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $lim_{x to 0} f(x) = lim_{x to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $lim_{x to a} f(x) neq L$ for some constant $L$.
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
add a comment |
up vote
0
down vote
express the first limit by the $epsilon/delta$ definition.
formally replace every occurrence of $x$ by $x-a$.
now you have the proof of the second limit.
I.e. rewrite
$$lim_{xto 0}f(x)=L
\iff
\forallepsilon>0:existsdelta>0:|x|<deltaimplies|f(x)-L|<epsilon
\iff
\forallepsilon>0:existsdelta>0:|x-a|<deltaimplies|f(x-a)-L|<epsilon
\iff\lim_{x-ato 0}f(x-a)=L.
$$
The argument holds because $x$ and $x-a$ cover the same set of values, and $lim_{x-ato0}$ is also $lim_{xto a}$.
answered Nov 23 at 17:41
Yves Daoust
123k668219
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2463895%2fepsilon-delta-proof-of-limits-being-equal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To show at least the second problem, assume that $lim_{xto 0} f(x) = L$. We will prove that $lim_{x to a}f(x-a) = L$.
By definition, we are to show : for all $epsilon > 0$, there exists $delta > 0$ such that
begin{equation}
|y - a |< delta implies |f(y-a) - L| < epsilon tag{1}
end{equation}
Fix one such $epsilon$, say $epsilon_0$.
We know that $lim_{x to 0} f(x) = L$. This means, that for the above $epsilon_0$, there exists some $delta_0 > 0$ such that begin{equation}|x - 0|( = |x|) < delta_0 implies |f(x) - L| < epsilon_0 tag{2}end{equation}
By taking $delta = delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
begin{equation}
|y-a| < delta_0
implies |f(y-a) - L |< epsilon_0 end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $lim_{x to 0} f(x) = lim_{x to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $lim_{x to a} f(x) neq L$ for some constant $L$.
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
add a comment |
up vote
1
down vote
accepted
To show at least the second problem, assume that $lim_{xto 0} f(x) = L$. We will prove that $lim_{x to a}f(x-a) = L$.
By definition, we are to show : for all $epsilon > 0$, there exists $delta > 0$ such that
begin{equation}
|y - a |< delta implies |f(y-a) - L| < epsilon tag{1}
end{equation}
Fix one such $epsilon$, say $epsilon_0$.
We know that $lim_{x to 0} f(x) = L$. This means, that for the above $epsilon_0$, there exists some $delta_0 > 0$ such that begin{equation}|x - 0|( = |x|) < delta_0 implies |f(x) - L| < epsilon_0 tag{2}end{equation}
By taking $delta = delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
begin{equation}
|y-a| < delta_0
implies |f(y-a) - L |< epsilon_0 end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $lim_{x to 0} f(x) = lim_{x to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $lim_{x to a} f(x) neq L$ for some constant $L$.
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To show at least the second problem, assume that $lim_{xto 0} f(x) = L$. We will prove that $lim_{x to a}f(x-a) = L$.
By definition, we are to show : for all $epsilon > 0$, there exists $delta > 0$ such that
begin{equation}
|y - a |< delta implies |f(y-a) - L| < epsilon tag{1}
end{equation}
Fix one such $epsilon$, say $epsilon_0$.
We know that $lim_{x to 0} f(x) = L$. This means, that for the above $epsilon_0$, there exists some $delta_0 > 0$ such that begin{equation}|x - 0|( = |x|) < delta_0 implies |f(x) - L| < epsilon_0 tag{2}end{equation}
By taking $delta = delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
begin{equation}
|y-a| < delta_0
implies |f(y-a) - L |< epsilon_0 end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $lim_{x to 0} f(x) = lim_{x to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $lim_{x to a} f(x) neq L$ for some constant $L$.
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
To show at least the second problem, assume that $lim_{xto 0} f(x) = L$. We will prove that $lim_{x to a}f(x-a) = L$.
By definition, we are to show : for all $epsilon > 0$, there exists $delta > 0$ such that
begin{equation}
|y - a |< delta implies |f(y-a) - L| < epsilon tag{1}
end{equation}
Fix one such $epsilon$, say $epsilon_0$.
We know that $lim_{x to 0} f(x) = L$. This means, that for the above $epsilon_0$, there exists some $delta_0 > 0$ such that begin{equation}|x - 0|( = |x|) < delta_0 implies |f(x) - L| < epsilon_0 tag{2}end{equation}
By taking $delta = delta_0$, we see that substituting $y-a = x$ in $(2)$ gives:
begin{equation}
|y-a| < delta_0
implies |f(y-a) - L |< epsilon_0 end{equation}
which is seen to be equivalent to statement $(1)$. Hence, $lim_{x to 0} f(x) = lim_{x to a} f(x-a)$ is true.
I urge you to prove that if one of these limits does not exist, then the other also doesn't. This involves negation of the definition of limit i.e. what does it mean for $lim_{x to a} f(x) neq L$ for some constant $L$.
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
edited Oct 9 '17 at 3:35
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
answered Oct 9 '17 at 3:22
астон вілла олоф мэллбэрг
37.2k33376
37.2k33376
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
add a comment |
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
Thank you! This was very clear and complete.
– Michael Connor
Oct 9 '17 at 13:43
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
You are welcome!
– астон вілла олоф мэллбэрг
Oct 9 '17 at 22:56
add a comment |
up vote
0
down vote
express the first limit by the $epsilon/delta$ definition.
formally replace every occurrence of $x$ by $x-a$.
now you have the proof of the second limit.
I.e. rewrite
$$lim_{xto 0}f(x)=L
\iff
\forallepsilon>0:existsdelta>0:|x|<deltaimplies|f(x)-L|<epsilon
\iff
\forallepsilon>0:existsdelta>0:|x-a|<deltaimplies|f(x-a)-L|<epsilon
\iff\lim_{x-ato 0}f(x-a)=L.
$$
The argument holds because $x$ and $x-a$ cover the same set of values, and $lim_{x-ato0}$ is also $lim_{xto a}$.
answered Nov 23 at 17:41
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
express the first limit by the $epsilon/delta$ definition.
formally replace every occurrence of $x$ by $x-a$.
now you have the proof of the second limit.
I.e. rewrite
$$lim_{xto 0}f(x)=L
\iff
\forallepsilon>0:existsdelta>0:|x|<deltaimplies|f(x)-L|<epsilon
\iff
\forallepsilon>0:existsdelta>0:|x-a|<deltaimplies|f(x-a)-L|<epsilon
\iff\lim_{x-ato 0}f(x-a)=L.
$$
The argument holds because $x$ and $x-a$ cover the same set of values, and $lim_{x-ato0}$ is also $lim_{xto a}$.
answered Nov 23 at 17:41
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
up vote
0
down vote
express the first limit by the $epsilon/delta$ definition.
formally replace every occurrence of $x$ by $x-a$.
now you have the proof of the second limit.
I.e. rewrite
$$lim_{xto 0}f(x)=L
\iff
\forallepsilon>0:existsdelta>0:|x|<deltaimplies|f(x)-L|<epsilon
\iff
\forallepsilon>0:existsdelta>0:|x-a|<deltaimplies|f(x-a)-L|<epsilon
\iff\lim_{x-ato 0}f(x-a)=L.
$$
The argument holds because $x$ and $x-a$ cover the same set of values, and $lim_{x-ato0}$ is also $lim_{xto a}$.
answered Nov 23 at 17:41
Yves Daoust
123k668219
express the first limit by the $epsilon/delta$ definition.
formally replace every occurrence of $x$ by $x-a$.
now you have the proof of the second limit.
I.e. rewrite
$$lim_{xto 0}f(x)=L
\iff
\forallepsilon>0:existsdelta>0:|x|<deltaimplies|f(x)-L|<epsilon
\iff
\forallepsilon>0:existsdelta>0:|x-a|<deltaimplies|f(x-a)-L|<epsilon
\iff\lim_{x-ato 0}f(x-a)=L.
$$
The argument holds because $x$ and $x-a$ cover the same set of values, and $lim_{x-ato0}$ is also $lim_{xto a}$.
answered Nov 23 at 17:41
Yves Daoust
123k668219
answered Nov 23 at 17:41
Yves Daoust
123k668219
answered Nov 23 at 17:41
Yves Daoust
123k668219
answered Nov 23 at 17:41
Yves Daoust
123k668219
123k668219
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2463895%2fepsilon-delta-proof-of-limits-being-equal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
If the two limits are unrelated, you have no other option than computing them (by the $epsilon/delta$ definition) and comparing the values.
– Yves Daoust
Nov 23 at 17:31