Proving function limit using $epsilon$-$delta$ definition
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
add a comment |
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
add a comment |
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
edited Dec 17 '18 at 16:14
Tianlalu
3,08621038
3,08621038
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
908
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
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%2f3044126%2fproving-function-limit-using-epsilon-delta-definition%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
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
edited Dec 17 '18 at 16:27
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
4059
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.
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%2f3044126%2fproving-function-limit-using-epsilon-delta-definition%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
