How can I find this limit? L'hopital's rule is never-ending
$begingroup$
I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.
I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?
Edit: would like to clarify I want to know how to find it algebraically
calculus limits
edited Jan 12 at 23:46
egreg
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
$endgroup$
add a comment |
$begingroup$
I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.
I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?
Edit: would like to clarify I want to know how to find it algebraically
calculus limits
edited Jan 12 at 23:46
egreg
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
$endgroup$
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25
add a comment |
$begingroup$
I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.
I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?
Edit: would like to clarify I want to know how to find it algebraically
calculus limits
edited Jan 12 at 23:46
egreg
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
$endgroup$
I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.
I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?
Edit: would like to clarify I want to know how to find it algebraically
calculus limits
calculus limits
edited Jan 12 at 23:46
egreg
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
edited Jan 12 at 23:46
egreg
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
edited Jan 12 at 23:46
egreg
186k1486208
edited Jan 12 at 23:46
egreg
186k1486208
edited Jan 12 at 23:46
egreg
186k1486208
186k1486208
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
asked Jan 12 at 23:18
Andrew WangAndrew Wang
415
415
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25
add a comment |
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$
answered Jan 12 at 23:23
pendermathpendermath
57612
$endgroup$
add a comment |
$begingroup$
Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
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%2f3071512%2fhow-can-i-find-this-limit-lhopitals-rule-is-never-ending%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
$begingroup$
You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$
answered Jan 12 at 23:23
pendermathpendermath
57612
$endgroup$
add a comment |
$begingroup$
You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$
answered Jan 12 at 23:23
pendermathpendermath
57612
$endgroup$
add a comment |
$begingroup$
You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$
answered Jan 12 at 23:23
pendermathpendermath
57612
$endgroup$
You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$
answered Jan 12 at 23:23
pendermathpendermath
57612
edited Jan 12 at 23:40
answered Jan 12 at 23:23
pendermathpendermath
57612
answered Jan 12 at 23:23
pendermathpendermath
57612
answered Jan 12 at 23:23
pendermathpendermath
57612
57612
add a comment |
add a comment |
$begingroup$
Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:58
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
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%2f3071512%2fhow-can-i-find-this-limit-lhopitals-rule-is-never-ending%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
$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53
$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25