Mixing
Find the limit of the sequence $frac{2^{n-1} + 3^n} {4^n - 1}$
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
edited Jan 8 at 8:47
E-mu
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
$endgroup$
add a comment |
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
edited Jan 8 at 8:47
E-mu
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
$endgroup$
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
edited Jan 8 at 8:47
E-mu
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
$endgroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
sequences-and-series limits exponential-function
edited Jan 8 at 8:47
E-mu
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
edited Jan 8 at 8:47
E-mu
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
edited Jan 8 at 8:47
E-mu
786417
edited Jan 8 at 8:47
E-mu
786417
edited Jan 8 at 8:47
E-mu
786417
786417
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
425
425
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
1
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
$endgroup$
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%2f3065938%2ffind-the-limit-of-the-sequence-frac2n-1-3n-4n-1%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
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
$endgroup$
add a comment |
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
$endgroup$
add a comment |
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
$endgroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
56.7k42159
56.7k42159
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%2f3065938%2ffind-the-limit-of-the-sequence-frac2n-1-3n-4n-1%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
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46