Mixing
Check my proof for this limit relation please
$begingroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
$endgroup$
add a comment |
$begingroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
$endgroup$
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
$begingroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
$endgroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
1065
1065
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
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',
autoActivateHeartbeat: false,
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%2f3066266%2fcheck-my-proof-for-this-limit-relation-please%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
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.
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%2f3066266%2fcheck-my-proof-for-this-limit-relation-please%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
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01