Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) over (4n^7 − 7n^2 + 5)} $ = 1/2
Multi tool use
Im trying to use the formal definition of a limit to prove
$lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$
$left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε
$ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7
$ {(1)over(ε+7)}$ < $4n^4$
$ {1over4ε+28^{1over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:
$ {1over4ε+28^{1over4}}$ < n
$ {1over4ε+28}$ < $n^4$
$ {1overε+7}$ < $4n^4$
$ {1over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with.
calculus limits analysis proof-writing
add a comment |
Im trying to use the formal definition of a limit to prove
$lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$
$left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε
$ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7
$ {(1)over(ε+7)}$ < $4n^4$
$ {1over4ε+28^{1over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:
$ {1over4ε+28^{1over4}}$ < n
$ {1over4ε+28}$ < $n^4$
$ {1overε+7}$ < $4n^4$
$ {1over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with.
calculus limits analysis proof-writing
add a comment |
Im trying to use the formal definition of a limit to prove
$lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$
$left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε
$ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7
$ {(1)over(ε+7)}$ < $4n^4$
$ {1over4ε+28^{1over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:
$ {1over4ε+28^{1over4}}$ < n
$ {1over4ε+28}$ < $n^4$
$ {1overε+7}$ < $4n^4$
$ {1over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with.
calculus limits analysis proof-writing
Im trying to use the formal definition of a limit to prove
$lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$
$left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε
$ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7
$ {(1)over(ε+7)}$ < $4n^4$
$ {1over4ε+28^{1over4}}$ < n = N
Now I have my N so I can writ my proof:
Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:
$ {1over4ε+28^{1over4}}$ < n
$ {1over4ε+28}$ < $n^4$
$ {1overε+7}$ < $4n^4$
$ {1over 4n^4}$ < $ε+7$
from there on I dont know how to use my inequalities to get back to what I started with.
calculus limits analysis proof-writing
calculus limits analysis proof-writing
asked Nov 26 '18 at 19:24
Malaikatu Kargbo
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}
add a comment |
I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
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%2f3014795%2fprove-the-limit-lim-n%25e2%2586%2592%25e2%2588%259e-2n7-3n5-4n-over-4n7-%25e2%2588%2592-7n2-5-1-2%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
Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}
add a comment |
Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}
add a comment |
Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}
Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}
answered Nov 26 '18 at 19:43
José Carlos Santos
150k22121221
150k22121221
add a comment |
add a comment |
I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
add a comment |
I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
add a comment |
I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$
I would rather use another approach:
Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
$$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$
answered Nov 26 '18 at 19:41
Dr. Mathva
919316
919316
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
add a comment |
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
– KM101
Nov 26 '18 at 19:49
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%2f3014795%2fprove-the-limit-lim-n%25e2%2586%2592%25e2%2588%259e-2n7-3n5-4n-over-4n7-%25e2%2588%2592-7n2-5-1-2%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
yxHJsUMADj,8 tPClK8ipZD6E,CGn4piXQoG