How do I compute $lim_{x to a}(a-x)tanleft(frac{πx}{2a}right)$
$begingroup$
Evaluate
$$lim_{x to a}f(x)=(a-x)tanleft(frac{πx}{2a}right)$$
I tried changing the tangent into cotangent by writing it in the form of $cotleft(dfrac{π}{2}-dfrac{πx}{2a}right)$. Then I factored out $dfrac{π}{2}$. But this didn't lead me to anything rigid. If anyone in the community could help me out, I'd really appreciate it.
limits limits-without-lhopital
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
$endgroup$
add a comment |
$begingroup$
Evaluate
$$lim_{x to a}f(x)=(a-x)tanleft(frac{πx}{2a}right)$$
I tried changing the tangent into cotangent by writing it in the form of $cotleft(dfrac{π}{2}-dfrac{πx}{2a}right)$. Then I factored out $dfrac{π}{2}$. But this didn't lead me to anything rigid. If anyone in the community could help me out, I'd really appreciate it.
limits limits-without-lhopital
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
$endgroup$
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25
add a comment |
$begingroup$
Evaluate
$$lim_{x to a}f(x)=(a-x)tanleft(frac{πx}{2a}right)$$
I tried changing the tangent into cotangent by writing it in the form of $cotleft(dfrac{π}{2}-dfrac{πx}{2a}right)$. Then I factored out $dfrac{π}{2}$. But this didn't lead me to anything rigid. If anyone in the community could help me out, I'd really appreciate it.
limits limits-without-lhopital
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
$endgroup$
Evaluate
$$lim_{x to a}f(x)=(a-x)tanleft(frac{πx}{2a}right)$$
I tried changing the tangent into cotangent by writing it in the form of $cotleft(dfrac{π}{2}-dfrac{πx}{2a}right)$. Then I factored out $dfrac{π}{2}$. But this didn't lead me to anything rigid. If anyone in the community could help me out, I'd really appreciate it.
limits limits-without-lhopital
limits limits-without-lhopital
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
edited Dec 11 '18 at 18:54
Lorenzo B.
1,8602520
1,8602520
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
asked Dec 11 '18 at 15:53
S.NepS.Nep
776
776
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25
add a comment |
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
As an alternative:
$$
lim_{x to a}f(x)=lim_{xto a}left((a-x)tanleft(frac{πx}{2a}right)right)
$$
Substitute $t = x - a iff x = t + a$, so your limit becomes:
$$
lim_{t to 0}f(t)=lim_{tto 0}left((a-(t+a))tanleft(frac{π(t+a)}{2a}right)right) = \
lim_{tto 0}left((-t)tanleft(frac{pi t + pi a}{2a}right)right) = -lim_{tto 0}left(ttanleft(frac{pi t}{2a} + frac{pi}{2}right)right) = \
= lim_{tto 0}left(tcotleft(frac{pi t}{2a}right)right) = lim_{tto0}frac{t}{tanleft(frac{pi t}{2a}right)}
$$
Now by Taylor of $tan x$ as $xto 0$:
$$
tan x sim x \
$$
We get:
$$
lim_{tto 0}frac{t}{tanleft(frac{pi t}{2a}right)} = lim_{tto 0}frac{2ta}{pi t} = fbox{$displaystyle frac{2a}{pi}$}
$$
answered Dec 13 '18 at 10:38
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
It does lead to a solution, keep going:
$$begin{align}lim_{x to a}(a-x)tanleft(dfrac{πx}{2a}right)&=lim_{x to a}(a-x)cotleft(dfrac{π}{2}left(1-frac xaright)right)=\
&=lim_{x to a}(a-x)cdot frac{color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}=\
&=lim_{x to a} color{red}{frac{dfrac{π}{2}left(1-frac xaright)}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}}cdot frac{2a}{pi}=\
&=frac{2a}{pi},end{align}$$
where it was used:
$$lim_{xto a} color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}=1;\
lim_{xto 0} color{red}{frac{x}{sin x}}=lim_{xto 0} frac{sin x}{x}=1.$$
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
Let $a-x=y$
For $ane0,$
$$tandfrac{pi x}{2a}=tandfrac{pi(a-y)}{2a}=cotdfrac{pi y}{2a}$$
Use $lim_{hto0}dfrac{sin h}h=1$
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$a not =0$.
Set $y =(πx)/(2a).$
Then $y rightarrow π/2.$
$((2a)/π)(π/2-y)dfrac{sin y}{cos y}.$
Recall $sin (π/2-y) =cos y.$
Then
$(2a/π)(dfrac {1}{dfrac{sin (π/2-y)}{π/2-y}})(sin y)$.
Can you take the limit $y rightarrow π/2?$
P.S. Recall $lim_{z rightarrow 0}dfrac{sin z}{z}=1$.
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
Your Answer
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%2f3035428%2fhow-do-i-compute-lim-x-to-aa-x-tan-left-frac%25cf%2580x2a-right%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
$begingroup$
As an alternative:
$$
lim_{x to a}f(x)=lim_{xto a}left((a-x)tanleft(frac{πx}{2a}right)right)
$$
Substitute $t = x - a iff x = t + a$, so your limit becomes:
$$
lim_{t to 0}f(t)=lim_{tto 0}left((a-(t+a))tanleft(frac{π(t+a)}{2a}right)right) = \
lim_{tto 0}left((-t)tanleft(frac{pi t + pi a}{2a}right)right) = -lim_{tto 0}left(ttanleft(frac{pi t}{2a} + frac{pi}{2}right)right) = \
= lim_{tto 0}left(tcotleft(frac{pi t}{2a}right)right) = lim_{tto0}frac{t}{tanleft(frac{pi t}{2a}right)}
$$
Now by Taylor of $tan x$ as $xto 0$:
$$
tan x sim x \
$$
We get:
$$
lim_{tto 0}frac{t}{tanleft(frac{pi t}{2a}right)} = lim_{tto 0}frac{2ta}{pi t} = fbox{$displaystyle frac{2a}{pi}$}
$$
answered Dec 13 '18 at 10:38
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
As an alternative:
$$
lim_{x to a}f(x)=lim_{xto a}left((a-x)tanleft(frac{πx}{2a}right)right)
$$
Substitute $t = x - a iff x = t + a$, so your limit becomes:
$$
lim_{t to 0}f(t)=lim_{tto 0}left((a-(t+a))tanleft(frac{π(t+a)}{2a}right)right) = \
lim_{tto 0}left((-t)tanleft(frac{pi t + pi a}{2a}right)right) = -lim_{tto 0}left(ttanleft(frac{pi t}{2a} + frac{pi}{2}right)right) = \
= lim_{tto 0}left(tcotleft(frac{pi t}{2a}right)right) = lim_{tto0}frac{t}{tanleft(frac{pi t}{2a}right)}
$$
Now by Taylor of $tan x$ as $xto 0$:
$$
tan x sim x \
$$
We get:
$$
lim_{tto 0}frac{t}{tanleft(frac{pi t}{2a}right)} = lim_{tto 0}frac{2ta}{pi t} = fbox{$displaystyle frac{2a}{pi}$}
$$
answered Dec 13 '18 at 10:38
romanroman
2,28421224
$endgroup$
add a comment |
$begingroup$
As an alternative:
$$
lim_{x to a}f(x)=lim_{xto a}left((a-x)tanleft(frac{πx}{2a}right)right)
$$
Substitute $t = x - a iff x = t + a$, so your limit becomes:
$$
lim_{t to 0}f(t)=lim_{tto 0}left((a-(t+a))tanleft(frac{π(t+a)}{2a}right)right) = \
lim_{tto 0}left((-t)tanleft(frac{pi t + pi a}{2a}right)right) = -lim_{tto 0}left(ttanleft(frac{pi t}{2a} + frac{pi}{2}right)right) = \
= lim_{tto 0}left(tcotleft(frac{pi t}{2a}right)right) = lim_{tto0}frac{t}{tanleft(frac{pi t}{2a}right)}
$$
Now by Taylor of $tan x$ as $xto 0$:
$$
tan x sim x \
$$
We get:
$$
lim_{tto 0}frac{t}{tanleft(frac{pi t}{2a}right)} = lim_{tto 0}frac{2ta}{pi t} = fbox{$displaystyle frac{2a}{pi}$}
$$
answered Dec 13 '18 at 10:38
romanroman
2,28421224
$endgroup$
As an alternative:
$$
lim_{x to a}f(x)=lim_{xto a}left((a-x)tanleft(frac{πx}{2a}right)right)
$$
Substitute $t = x - a iff x = t + a$, so your limit becomes:
$$
lim_{t to 0}f(t)=lim_{tto 0}left((a-(t+a))tanleft(frac{π(t+a)}{2a}right)right) = \
lim_{tto 0}left((-t)tanleft(frac{pi t + pi a}{2a}right)right) = -lim_{tto 0}left(ttanleft(frac{pi t}{2a} + frac{pi}{2}right)right) = \
= lim_{tto 0}left(tcotleft(frac{pi t}{2a}right)right) = lim_{tto0}frac{t}{tanleft(frac{pi t}{2a}right)}
$$
Now by Taylor of $tan x$ as $xto 0$:
$$
tan x sim x \
$$
We get:
$$
lim_{tto 0}frac{t}{tanleft(frac{pi t}{2a}right)} = lim_{tto 0}frac{2ta}{pi t} = fbox{$displaystyle frac{2a}{pi}$}
$$
answered Dec 13 '18 at 10:38
romanroman
2,28421224
answered Dec 13 '18 at 10:38
romanroman
2,28421224
answered Dec 13 '18 at 10:38
romanroman
2,28421224
answered Dec 13 '18 at 10:38
romanroman
2,28421224
2,28421224
add a comment |
add a comment |
$begingroup$
It does lead to a solution, keep going:
$$begin{align}lim_{x to a}(a-x)tanleft(dfrac{πx}{2a}right)&=lim_{x to a}(a-x)cotleft(dfrac{π}{2}left(1-frac xaright)right)=\
&=lim_{x to a}(a-x)cdot frac{color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}=\
&=lim_{x to a} color{red}{frac{dfrac{π}{2}left(1-frac xaright)}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}}cdot frac{2a}{pi}=\
&=frac{2a}{pi},end{align}$$
where it was used:
$$lim_{xto a} color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}=1;\
lim_{xto 0} color{red}{frac{x}{sin x}}=lim_{xto 0} frac{sin x}{x}=1.$$
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
It does lead to a solution, keep going:
$$begin{align}lim_{x to a}(a-x)tanleft(dfrac{πx}{2a}right)&=lim_{x to a}(a-x)cotleft(dfrac{π}{2}left(1-frac xaright)right)=\
&=lim_{x to a}(a-x)cdot frac{color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}=\
&=lim_{x to a} color{red}{frac{dfrac{π}{2}left(1-frac xaright)}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}}cdot frac{2a}{pi}=\
&=frac{2a}{pi},end{align}$$
where it was used:
$$lim_{xto a} color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}=1;\
lim_{xto 0} color{red}{frac{x}{sin x}}=lim_{xto 0} frac{sin x}{x}=1.$$
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
It does lead to a solution, keep going:
$$begin{align}lim_{x to a}(a-x)tanleft(dfrac{πx}{2a}right)&=lim_{x to a}(a-x)cotleft(dfrac{π}{2}left(1-frac xaright)right)=\
&=lim_{x to a}(a-x)cdot frac{color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}=\
&=lim_{x to a} color{red}{frac{dfrac{π}{2}left(1-frac xaright)}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}}cdot frac{2a}{pi}=\
&=frac{2a}{pi},end{align}$$
where it was used:
$$lim_{xto a} color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}=1;\
lim_{xto 0} color{red}{frac{x}{sin x}}=lim_{xto 0} frac{sin x}{x}=1.$$
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
$endgroup$
It does lead to a solution, keep going:
$$begin{align}lim_{x to a}(a-x)tanleft(dfrac{πx}{2a}right)&=lim_{x to a}(a-x)cotleft(dfrac{π}{2}left(1-frac xaright)right)=\
&=lim_{x to a}(a-x)cdot frac{color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}=\
&=lim_{x to a} color{red}{frac{dfrac{π}{2}left(1-frac xaright)}{sinleft(dfrac{π}{2}left(1-frac xaright)right)}}cdot frac{2a}{pi}=\
&=frac{2a}{pi},end{align}$$
where it was used:
$$lim_{xto a} color{blue}{cosleft(dfrac{π}{2}left(1-frac xaright)right)}=1;\
lim_{xto 0} color{red}{frac{x}{sin x}}=lim_{xto 0} frac{sin x}{x}=1.$$
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:26
farruhotafarruhota
20.5k2739
20.5k2739
add a comment |
add a comment |
$begingroup$
Let $a-x=y$
For $ane0,$
$$tandfrac{pi x}{2a}=tandfrac{pi(a-y)}{2a}=cotdfrac{pi y}{2a}$$
Use $lim_{hto0}dfrac{sin h}h=1$
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Let $a-x=y$
For $ane0,$
$$tandfrac{pi x}{2a}=tandfrac{pi(a-y)}{2a}=cotdfrac{pi y}{2a}$$
Use $lim_{hto0}dfrac{sin h}h=1$
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Let $a-x=y$
For $ane0,$
$$tandfrac{pi x}{2a}=tandfrac{pi(a-y)}{2a}=cotdfrac{pi y}{2a}$$
Use $lim_{hto0}dfrac{sin h}h=1$
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Let $a-x=y$
For $ane0,$
$$tandfrac{pi x}{2a}=tandfrac{pi(a-y)}{2a}=cotdfrac{pi y}{2a}$$
Use $lim_{hto0}dfrac{sin h}h=1$
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 11 '18 at 15:58
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
$a not =0$.
Set $y =(πx)/(2a).$
Then $y rightarrow π/2.$
$((2a)/π)(π/2-y)dfrac{sin y}{cos y}.$
Recall $sin (π/2-y) =cos y.$
Then
$(2a/π)(dfrac {1}{dfrac{sin (π/2-y)}{π/2-y}})(sin y)$.
Can you take the limit $y rightarrow π/2?$
P.S. Recall $lim_{z rightarrow 0}dfrac{sin z}{z}=1$.
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$a not =0$.
Set $y =(πx)/(2a).$
Then $y rightarrow π/2.$
$((2a)/π)(π/2-y)dfrac{sin y}{cos y}.$
Recall $sin (π/2-y) =cos y.$
Then
$(2a/π)(dfrac {1}{dfrac{sin (π/2-y)}{π/2-y}})(sin y)$.
Can you take the limit $y rightarrow π/2?$
P.S. Recall $lim_{z rightarrow 0}dfrac{sin z}{z}=1$.
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$a not =0$.
Set $y =(πx)/(2a).$
Then $y rightarrow π/2.$
$((2a)/π)(π/2-y)dfrac{sin y}{cos y}.$
Recall $sin (π/2-y) =cos y.$
Then
$(2a/π)(dfrac {1}{dfrac{sin (π/2-y)}{π/2-y}})(sin y)$.
Can you take the limit $y rightarrow π/2?$
P.S. Recall $lim_{z rightarrow 0}dfrac{sin z}{z}=1$.
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$a not =0$.
Set $y =(πx)/(2a).$
Then $y rightarrow π/2.$
$((2a)/π)(π/2-y)dfrac{sin y}{cos y}.$
Recall $sin (π/2-y) =cos y.$
Then
$(2a/π)(dfrac {1}{dfrac{sin (π/2-y)}{π/2-y}})(sin y)$.
Can you take the limit $y rightarrow π/2?$
P.S. Recall $lim_{z rightarrow 0}dfrac{sin z}{z}=1$.
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
edited Dec 13 '18 at 8:49
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
answered Dec 11 '18 at 16:25
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
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%2f3035428%2fhow-do-i-compute-lim-x-to-aa-x-tan-left-frac%25cf%2580x2a-right%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
$begingroup$
Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour
$endgroup$
– Robert Z
Dec 12 '18 at 16:25