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 $limlimits_xto infty (x arctan x - fracxpi2)$
Airplane gently rocking its wings during whole flight
My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?
Can you teleport closer to a creature you are Frightened of?
Reshaping json / reparing json inside shell script (remove trailing comma)
What was Carter Burke's job for "the company" in Aliens?
It is correct to match light sources with the same color temperature?
Cannot shrink btrfs filesystem although there is still data and metadata space left : ERROR: unable to resize '/home': No space left on device
Ising model simulation
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what chip
Is there a difference between "Fahrstuhl" and "Aufzug"?
What does "shotgun unity" refer to here in this sentence?
Asymptote: 3d graph over a disc
Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?
Where do students learn to solve polynomial equations these days?
Lucky Feat: How can "more than one creature spend a luck point to influence the outcome of a roll"?
(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?
How to avoid supervisors with prejudiced views?
What's the commands of Cisco query bgp neighbor table, bgp table and router table?
Do scriptures give a method to recognize a truly self-realized person/jivanmukta?
What CSS properties can the br tag have?
From jafe to El-Guest
Are the names of these months realistic?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Man transported from Alternate World into ours by a Neutrino Detector
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
1,8411823
$endgroup$
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
1,8411823
$endgroup$
$begingroup$
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
1,8411823
$endgroup$
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
1,8411823
asked 2 hours ago
lanse7ptylanse7pty
1,8411823
edited 1 hour ago
lanse7pty
asked 2 hours ago
lanse7ptylanse7pty
1,8411823
asked 2 hours ago
lanse7ptylanse7pty
1,8411823
asked 2 hours ago
lanse7ptylanse7pty
1,8411823
1,8411823
$begingroup$
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 |
4 Answers
4
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 2 hours ago
Paras KhoslaParas Khosla
2,736423
$endgroup$
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
285
$endgroup$
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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
4 Answers
4
active
oldest
votes
4 Answers
4
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 2 hours ago
Paras KhoslaParas Khosla
2,736423
$endgroup$
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 2 hours ago
Paras KhoslaParas Khosla
2,736423
$endgroup$
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 2 hours ago
Paras KhoslaParas Khosla
2,736423
$endgroup$
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 2 hours ago
Paras KhoslaParas Khosla
2,736423
edited 1 hour ago
answered 2 hours ago
Paras KhoslaParas Khosla
2,736423
answered 2 hours ago
Paras KhoslaParas Khosla
2,736423
answered 2 hours ago
Paras KhoslaParas Khosla
2,736423
2,736423
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
285
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
285
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
285
$endgroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
285
answered 2 hours ago
AdmuthAdmuth
285
answered 2 hours ago
AdmuthAdmuth
285
answered 2 hours ago
AdmuthAdmuth
285
285
add a comment |
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
$endgroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
answered 9 mins ago
trancelocationtrancelocation
13.4k1827
13.4k1827
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