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 $limlimits_xto infty (x arctan x - fracxpi2)$
How to calculate the two limits?
Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact
Can I board the first leg of the flight without having final country's visa?
Is a distribution that is normal, but highly skewed, considered Gaussian?
IC has pull-down resistors on SMBus lines?
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
How do I fit a non linear curve?
What are the unusually-enlarged wing sections on this P-38 Lightning?
What is the difference between "hamstring tendon" and "common hamstring tendon"?
Lucky Feat: How can "more than one creature spend a luck point to influence the outcome of a roll"?
Won the lottery - how do I keep the money?
What is the process for purifying your home if you believe it may have been previously used for pagan worship?
Is there an equivalent of cd - for cp or mv
Vector calculus integration identity problem
Which Pokemon have a special animation when running with them out of their pokeball?
Aggressive Under-Indexing and no data for missing index
Towers in the ocean; How deep can they be built?
Man transported from Alternate World into ours by a Neutrino Detector
Is there such a thing as a proper verb, like a proper noun?
Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?
Reshaping json / reparing json inside shell script (remove trailing comma)
Plausibility of squid whales
What difference does it make using sed with/without whitespaces?
How to use ReplaceAll on an expression that contains a rule
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 2 hours 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
385
$endgroup$
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 13 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
385
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
385
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
385
$endgroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
385
answered 2 hours ago
AdmuthAdmuth
385
answered 2 hours ago
AdmuthAdmuth
385
answered 2 hours ago
AdmuthAdmuth
385
385
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 13 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 13 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 13 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 13 mins ago
trancelocationtrancelocation
13.4k1827
answered 13 mins ago
trancelocationtrancelocation
13.4k1827
answered 13 mins ago
trancelocationtrancelocation
13.4k1827
answered 13 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