Mixing
How to find the limit $lim_{ntoinfty} {2^n+(-1)^n over 2^{n+1}+(-1)^{n+1}}$?
$begingroup$
How do I find this limit?
$$lim_{n to infty} {2^n+(-1)^n over 2^{n+1}+(-1)^{n+1}}$$
I don't know how to take the limit of $lim_{n to infty}{(-1)^n}$, I know that the limit $lim_{n to infty} {2^n}$ and $lim_{n to infty} {2^{n+1}}$ tend both to infinity.
I asked my teacher and he said that the solution was ${1 over 2}$ but I don't know why, I tried to solve it but I got undefined as the answer, how do I solve this correctly?
When I did was $$lim_{n to infty} {1+{(-1)^nover2^n} over 2+{(-1)^{n+1}over2^n}}$$ I took the limit of that, and ended up like this:
$$ {1+lim_{n to infty}{(-1)^nover2^n} over 2+lim_{n to infty}{(-1)^{n+1}over2^n}}$$ But both $lim_{n to infty}{(-1)^{n}over2^n}$ and $lim_{n to infty}{(-1)^{n+1}over2^n}$ gives me undefined, So I eneded up with $1+undefinedover2+undefined$
calculus sequences-and-series limits
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
$endgroup$
|
show 4 more comments
$begingroup$
How do I find this limit?
$$lim_{n to infty} {2^n+(-1)^n over 2^{n+1}+(-1)^{n+1}}$$
I don't know how to take the limit of $lim_{n to infty}{(-1)^n}$, I know that the limit $lim_{n to infty} {2^n}$ and $lim_{n to infty} {2^{n+1}}$ tend both to infinity.
I asked my teacher and he said that the solution was ${1 over 2}$ but I don't know why, I tried to solve it but I got undefined as the answer, how do I solve this correctly?
When I did was $$lim_{n to infty} {1+{(-1)^nover2^n} over 2+{(-1)^{n+1}over2^n}}$$ I took the limit of that, and ended up like this:
$$ {1+lim_{n to infty}{(-1)^nover2^n} over 2+lim_{n to infty}{(-1)^{n+1}over2^n}}$$ But both $lim_{n to infty}{(-1)^{n}over2^n}$ and $lim_{n to infty}{(-1)^{n+1}over2^n}$ gives me undefined, So I eneded up with $1+undefinedover2+undefined$
calculus sequences-and-series limits
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
$endgroup$
1
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13
|
show 4 more comments
$begingroup$
How do I find this limit?
$$lim_{n to infty} {2^n+(-1)^n over 2^{n+1}+(-1)^{n+1}}$$
I don't know how to take the limit of $lim_{n to infty}{(-1)^n}$, I know that the limit $lim_{n to infty} {2^n}$ and $lim_{n to infty} {2^{n+1}}$ tend both to infinity.
I asked my teacher and he said that the solution was ${1 over 2}$ but I don't know why, I tried to solve it but I got undefined as the answer, how do I solve this correctly?
When I did was $$lim_{n to infty} {1+{(-1)^nover2^n} over 2+{(-1)^{n+1}over2^n}}$$ I took the limit of that, and ended up like this:
$$ {1+lim_{n to infty}{(-1)^nover2^n} over 2+lim_{n to infty}{(-1)^{n+1}over2^n}}$$ But both $lim_{n to infty}{(-1)^{n}over2^n}$ and $lim_{n to infty}{(-1)^{n+1}over2^n}$ gives me undefined, So I eneded up with $1+undefinedover2+undefined$
calculus sequences-and-series limits
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
$endgroup$
How do I find this limit?
$$lim_{n to infty} {2^n+(-1)^n over 2^{n+1}+(-1)^{n+1}}$$
I don't know how to take the limit of $lim_{n to infty}{(-1)^n}$, I know that the limit $lim_{n to infty} {2^n}$ and $lim_{n to infty} {2^{n+1}}$ tend both to infinity.
I asked my teacher and he said that the solution was ${1 over 2}$ but I don't know why, I tried to solve it but I got undefined as the answer, how do I solve this correctly?
When I did was $$lim_{n to infty} {1+{(-1)^nover2^n} over 2+{(-1)^{n+1}over2^n}}$$ I took the limit of that, and ended up like this:
$$ {1+lim_{n to infty}{(-1)^nover2^n} over 2+lim_{n to infty}{(-1)^{n+1}over2^n}}$$ But both $lim_{n to infty}{(-1)^{n}over2^n}$ and $lim_{n to infty}{(-1)^{n+1}over2^n}$ gives me undefined, So I eneded up with $1+undefinedover2+undefined$
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
edited Jan 23 at 3:17
davidllerenav
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
asked Jan 22 at 1:30
davidllerenavdavidllerenav
1838
1838
1
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13
|
show 4 more comments
1
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13
1
1
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If you divide the top and bottom by $2^n$, you get
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}}$$
Now seeing that both $frac{{(-1)}^n}{2^n}$ and $frac{{(-1)}^{n+1}}{2^n}$ approach $0$ as $xtoinfty$ (gotta use some intuition here), we get that
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}} = frac{1+0}{2+0} = frac{1}{2}$$
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
$endgroup$
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
|
show 2 more comments
$begingroup$
$
lim_{ntoinfty} frac{ 2^n+(-1)^n }{ 2^{n+1}+(-1)^{n+1} } \
= lim_{ntoinfty} frac{ frac{2^n}{2^n} + frac{{(-1)}^n}{2^n} }{ frac{2^{n+1}}{2^n} + frac{{(-1)}^{n+1}}{2^n} } \
= lim_{ntoinfty} frac{ 1 + (frac{-1}{2})^n }{ 2 - (frac{-1}{2})^n } \
= frac{ 1 + lim_{ntoinfty} (frac{-1}{2})^n }{ 2 - lim_{ntoinfty} (frac{-1}{2})^n } \ $
Using Squeeze theorem, and since $
-(frac{1}{2})^n le (frac{-1}{2})^n le (frac{1}{2})^n forall n ge 1 $ and $
lim_{ntoinfty} -(frac{1}{2})^n = lim_{ntoinfty} (frac{1}{2})^n = 0 $ , we get
$
= frac{ 1 + 0 }{ 2 - 0 } [ because lim_{ntoinfty} (frac{-1}{2})^n = 0 ] \
= frac{1}{2}
$
answered Jan 22 at 7:15
programmerprogrammer
816
$endgroup$
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%2f3082652%2fhow-to-find-the-limit-lim-n-to-infty-2n-1n-over-2n1-1n1%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
$begingroup$
If you divide the top and bottom by $2^n$, you get
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}}$$
Now seeing that both $frac{{(-1)}^n}{2^n}$ and $frac{{(-1)}^{n+1}}{2^n}$ approach $0$ as $xtoinfty$ (gotta use some intuition here), we get that
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}} = frac{1+0}{2+0} = frac{1}{2}$$
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
$endgroup$
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
|
show 2 more comments
$begingroup$
If you divide the top and bottom by $2^n$, you get
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}}$$
Now seeing that both $frac{{(-1)}^n}{2^n}$ and $frac{{(-1)}^{n+1}}{2^n}$ approach $0$ as $xtoinfty$ (gotta use some intuition here), we get that
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}} = frac{1+0}{2+0} = frac{1}{2}$$
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
$endgroup$
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
|
show 2 more comments
$begingroup$
If you divide the top and bottom by $2^n$, you get
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}}$$
Now seeing that both $frac{{(-1)}^n}{2^n}$ and $frac{{(-1)}^{n+1}}{2^n}$ approach $0$ as $xtoinfty$ (gotta use some intuition here), we get that
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}} = frac{1+0}{2+0} = frac{1}{2}$$
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
$endgroup$
If you divide the top and bottom by $2^n$, you get
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}}$$
Now seeing that both $frac{{(-1)}^n}{2^n}$ and $frac{{(-1)}^{n+1}}{2^n}$ approach $0$ as $xtoinfty$ (gotta use some intuition here), we get that
$$lim_{ntoinfty} frac{1 + frac{{(-1)}^n}{2^n}}{2 + frac{{(-1)}^{n+1}}{2^n}} = frac{1+0}{2+0} = frac{1}{2}$$
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
edited Jan 22 at 2:30
J. W. Tanner
2,9191217
2,9191217
answered Jan 22 at 1:59
HyperionHyperion
636110
answered Jan 22 at 1:59
HyperionHyperion
636110
answered Jan 22 at 1:59
HyperionHyperion
636110
636110
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
|
show 2 more comments
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
But wouldn't $(-1)^n$ and $(-1)^{n+1}$ diverge? Because it is -1 and 1 if n is even or odd.
$endgroup$
– davidllerenav
Jan 22 at 2:08
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
@davidllerenav We can write the two fractional sequences as $(-frac{1}{2})^n$ and $-(-frac{1}{2})^n$. Even though these sequences "bounce" back and forth between negative and positive, they still approach $0$.
$endgroup$
– Hyperion
Jan 22 at 2:11
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
Maybe you could make it more rigorous with the squeeze theorem
$endgroup$
– J. W. Tanner
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@Hyperion That is because we write it like $({-1over2})^n$ and $({-1over2})^{n+1}$, right? Because I saw on my book that the $lim_{ntoinfty} a^n=0$ if |a|<1.
$endgroup$
– davidllerenav
Jan 22 at 2:21
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
$begingroup$
@davidllerenav Watch out on on your second fraction; $(-frac{1}{2})^{n+1} neq -(-frac{1}{2})^{n}$, but other than that, your reasoning is correct.
$endgroup$
– Hyperion
Jan 22 at 2:33
|
show 2 more comments
$begingroup$
$
lim_{ntoinfty} frac{ 2^n+(-1)^n }{ 2^{n+1}+(-1)^{n+1} } \
= lim_{ntoinfty} frac{ frac{2^n}{2^n} + frac{{(-1)}^n}{2^n} }{ frac{2^{n+1}}{2^n} + frac{{(-1)}^{n+1}}{2^n} } \
= lim_{ntoinfty} frac{ 1 + (frac{-1}{2})^n }{ 2 - (frac{-1}{2})^n } \
= frac{ 1 + lim_{ntoinfty} (frac{-1}{2})^n }{ 2 - lim_{ntoinfty} (frac{-1}{2})^n } \ $
Using Squeeze theorem, and since $
-(frac{1}{2})^n le (frac{-1}{2})^n le (frac{1}{2})^n forall n ge 1 $ and $
lim_{ntoinfty} -(frac{1}{2})^n = lim_{ntoinfty} (frac{1}{2})^n = 0 $ , we get
$
= frac{ 1 + 0 }{ 2 - 0 } [ because lim_{ntoinfty} (frac{-1}{2})^n = 0 ] \
= frac{1}{2}
$
answered Jan 22 at 7:15
programmerprogrammer
816
$endgroup$
add a comment |
$begingroup$
$
lim_{ntoinfty} frac{ 2^n+(-1)^n }{ 2^{n+1}+(-1)^{n+1} } \
= lim_{ntoinfty} frac{ frac{2^n}{2^n} + frac{{(-1)}^n}{2^n} }{ frac{2^{n+1}}{2^n} + frac{{(-1)}^{n+1}}{2^n} } \
= lim_{ntoinfty} frac{ 1 + (frac{-1}{2})^n }{ 2 - (frac{-1}{2})^n } \
= frac{ 1 + lim_{ntoinfty} (frac{-1}{2})^n }{ 2 - lim_{ntoinfty} (frac{-1}{2})^n } \ $
Using Squeeze theorem, and since $
-(frac{1}{2})^n le (frac{-1}{2})^n le (frac{1}{2})^n forall n ge 1 $ and $
lim_{ntoinfty} -(frac{1}{2})^n = lim_{ntoinfty} (frac{1}{2})^n = 0 $ , we get
$
= frac{ 1 + 0 }{ 2 - 0 } [ because lim_{ntoinfty} (frac{-1}{2})^n = 0 ] \
= frac{1}{2}
$
answered Jan 22 at 7:15
programmerprogrammer
816
$endgroup$
add a comment |
$begingroup$
$
lim_{ntoinfty} frac{ 2^n+(-1)^n }{ 2^{n+1}+(-1)^{n+1} } \
= lim_{ntoinfty} frac{ frac{2^n}{2^n} + frac{{(-1)}^n}{2^n} }{ frac{2^{n+1}}{2^n} + frac{{(-1)}^{n+1}}{2^n} } \
= lim_{ntoinfty} frac{ 1 + (frac{-1}{2})^n }{ 2 - (frac{-1}{2})^n } \
= frac{ 1 + lim_{ntoinfty} (frac{-1}{2})^n }{ 2 - lim_{ntoinfty} (frac{-1}{2})^n } \ $
Using Squeeze theorem, and since $
-(frac{1}{2})^n le (frac{-1}{2})^n le (frac{1}{2})^n forall n ge 1 $ and $
lim_{ntoinfty} -(frac{1}{2})^n = lim_{ntoinfty} (frac{1}{2})^n = 0 $ , we get
$
= frac{ 1 + 0 }{ 2 - 0 } [ because lim_{ntoinfty} (frac{-1}{2})^n = 0 ] \
= frac{1}{2}
$
answered Jan 22 at 7:15
programmerprogrammer
816
$endgroup$
$
lim_{ntoinfty} frac{ 2^n+(-1)^n }{ 2^{n+1}+(-1)^{n+1} } \
= lim_{ntoinfty} frac{ frac{2^n}{2^n} + frac{{(-1)}^n}{2^n} }{ frac{2^{n+1}}{2^n} + frac{{(-1)}^{n+1}}{2^n} } \
= lim_{ntoinfty} frac{ 1 + (frac{-1}{2})^n }{ 2 - (frac{-1}{2})^n } \
= frac{ 1 + lim_{ntoinfty} (frac{-1}{2})^n }{ 2 - lim_{ntoinfty} (frac{-1}{2})^n } \ $
Using Squeeze theorem, and since $
-(frac{1}{2})^n le (frac{-1}{2})^n le (frac{1}{2})^n forall n ge 1 $ and $
lim_{ntoinfty} -(frac{1}{2})^n = lim_{ntoinfty} (frac{1}{2})^n = 0 $ , we get
$
= frac{ 1 + 0 }{ 2 - 0 } [ because lim_{ntoinfty} (frac{-1}{2})^n = 0 ] \
= frac{1}{2}
$
answered Jan 22 at 7:15
programmerprogrammer
816
answered Jan 22 at 7:15
programmerprogrammer
816
answered Jan 22 at 7:15
programmerprogrammer
816
answered Jan 22 at 7:15
programmerprogrammer
816
816
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%2f3082652%2fhow-to-find-the-limit-lim-n-to-infty-2n-1n-over-2n1-1n1%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
$begingroup$
Try dividing top and bottom by $2^n$
$endgroup$
– J. W. Tanner
Jan 22 at 1:32
$begingroup$
Then I get $lim_{ntoinfty} {1+{(-1)^nover 2^n}over2+{(-1)^{n+1}over 2^n}}$ , but what happens with ${(-1)^nover 2^n}$ and ${(-1)^{n+1}over 2^n}$?
$endgroup$
– davidllerenav
Jan 22 at 1:45
$begingroup$
Do you know $ lim_{n to infty} {frac {1 + {1 over 2^n }} {2 + {1 over 2^n}}}$ ?
$endgroup$
– J. W. Tanner
Jan 22 at 2:02
$begingroup$
I think it would be ${1 over 2}$, right? Because $lim_{ntoinfty} {1over2^n}$ is 0.
$endgroup$
– davidllerenav
Jan 22 at 2:06
$begingroup$
Right, it would be $1 over 2$
$endgroup$
– J. W. Tanner
Jan 22 at 2:13