How to calculate the two limits? The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$
Players Circumventing the limitations of Wish
Calculate the Mean mean of two numbers
How do I fit a non linear curve?
How to get the last not-null value in an ordered column of a huge table?
Cannot shrink btrfs filesystem although there is still data and metadata space left : ERROR: unable to resize '/home': No space left on device
What are the unusually-enlarged wing sections on this P-38 Lightning?
(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?
What was Carter Burke's job for "the company" in Aliens?
Physiological effects of huge anime eyes
What does "shotgun unity" refer to here in this sentence?
Is it OK to decorate a log book cover?
Towers in the ocean; How deep can they be built?
Is French Guiana a (hard) EU border?
Can you teleport closer to a creature you are Frightened of?
Does higher Oxidation/ reduction potential translate to higher energy storage in battery?
Plausibility of squid whales
Computationally populating tables with probability data
Spaces in which all closed sets are regular closed
Is there a way to save my career from absolute disaster?
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what chip
Inductor and Capacitor in Parallel
Reshaping json / reparing json inside shell script (remove trailing comma)
Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank
What connection does MS Office have to Netscape Navigator?
How to calculate the two limits?
The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
add a comment |
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago
add a comment |
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
limits
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
edited 1 hour ago
lanse7pty
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
1,8361823
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago
add a comment |
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
add a comment |
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%2f3170200%2fhow-to-calculate-the-two-limits%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
edited 1 hour ago
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
answered 1 hour ago
Paras KhoslaParas Khosla
2,726423
2,726423
add a comment |
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
add a comment |
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
185
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%2f3170200%2fhow-to-calculate-the-two-limits%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
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago