Mixing
$lim_{n to infty}frac{e^sqrt{n}}{c^n}$?
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
add a comment |
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
asked Feb 3 at 12:09
doobledooble
153
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
asked Feb 3 at 12:09
doobledooble
153
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
edited Feb 3 at 12:53
Thomas Shelby
4,7382727
4,7382727
asked Feb 3 at 12:09
doobledooble
153
asked Feb 3 at 12:09
doobledooble
153
asked Feb 3 at 12:09
doobledooble
153
153
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
Your Answer
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%2f3098499%2flim-n-to-infty-frace-sqrtncn%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
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
45118
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%2f3098499%2flim-n-to-infty-frace-sqrtncn%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
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16