How to prove that the limit of this equation is 400/pi

Clash Royale CLAN TAG#URR8PPP

up vote
3
down vote

favorite

I am new to this forum, so I hope I am proceeding in the correct way. Please excuse any mistakes.

I am trying to prove that the limiting factor of a 2D shape's surface area is a circle, and I have managed to find an equation for this from a regular polygon. In my question, I assume that the maximum perimeter/circumference one can form is 40cm (i.e. if someone has 40cm of string). So:

For a perimeter of 40:
$$
lim_n rightarrow inftyfrac200sin(frac2πn)nsin^2(fracπn)=frac400π
$$

I got this value using Wolfram Alpha, which confirmed that the limiting factor for surface area is a circle, since the surface area of a circle with a circumference of 40cm = 400/π.

However, I am unable to prove this formula algebraically. I tried using l'hospital's rule after realising that the limit of the original function was 0/0, but I got nowhere. In fact, the result I got was -400π/0, which was very disappointing after so much working out!

I was wondering if anyone could help me prove this algebraically or otherwise. I am happy that I found an equation that proves what I wanted to, but I am unable to prove Wolfram Alpha's result, which is frustrating.

Again, I am new to this forum, so please let me know if I have made any mistakes so that I can edit my question.

calculus limits limits-without-lhopital

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
  – jacob1729
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
  – Eric Spies
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

I am new to this forum, so I hope I am proceeding in the correct way. Please excuse any mistakes.

I am trying to prove that the limiting factor of a 2D shape's surface area is a circle, and I have managed to find an equation for this from a regular polygon. In my question, I assume that the maximum perimeter/circumference one can form is 40cm (i.e. if someone has 40cm of string). So:

For a perimeter of 40:
$$
lim_n rightarrow inftyfrac200sin(frac2πn)nsin^2(fracπn)=frac400π
$$

I got this value using Wolfram Alpha, which confirmed that the limiting factor for surface area is a circle, since the surface area of a circle with a circumference of 40cm = 400/π.

However, I am unable to prove this formula algebraically. I tried using l'hospital's rule after realising that the limit of the original function was 0/0, but I got nowhere. In fact, the result I got was -400π/0, which was very disappointing after so much working out!

I was wondering if anyone could help me prove this algebraically or otherwise. I am happy that I found an equation that proves what I wanted to, but I am unable to prove Wolfram Alpha's result, which is frustrating.

Again, I am new to this forum, so please let me know if I have made any mistakes so that I can edit my question.

calculus limits limits-without-lhopital

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
  – jacob1729
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
  – Eric Spies
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I am new to this forum, so I hope I am proceeding in the correct way. Please excuse any mistakes.

I am trying to prove that the limiting factor of a 2D shape's surface area is a circle, and I have managed to find an equation for this from a regular polygon. In my question, I assume that the maximum perimeter/circumference one can form is 40cm (i.e. if someone has 40cm of string). So:

For a perimeter of 40:
$$
lim_n rightarrow inftyfrac200sin(frac2πn)nsin^2(fracπn)=frac400π
$$

I got this value using Wolfram Alpha, which confirmed that the limiting factor for surface area is a circle, since the surface area of a circle with a circumference of 40cm = 400/π.

However, I am unable to prove this formula algebraically. I tried using l'hospital's rule after realising that the limit of the original function was 0/0, but I got nowhere. In fact, the result I got was -400π/0, which was very disappointing after so much working out!

I was wondering if anyone could help me prove this algebraically or otherwise. I am happy that I found an equation that proves what I wanted to, but I am unable to prove Wolfram Alpha's result, which is frustrating.

Again, I am new to this forum, so please let me know if I have made any mistakes so that I can edit my question.

calculus limits limits-without-lhopital

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I am new to this forum, so I hope I am proceeding in the correct way. Please excuse any mistakes.

I am trying to prove that the limiting factor of a 2D shape's surface area is a circle, and I have managed to find an equation for this from a regular polygon. In my question, I assume that the maximum perimeter/circumference one can form is 40cm (i.e. if someone has 40cm of string). So:

For a perimeter of 40:
$$
lim_n rightarrow inftyfrac200sin(frac2πn)nsin^2(fracπn)=frac400π
$$

I got this value using Wolfram Alpha, which confirmed that the limiting factor for surface area is a circle, since the surface area of a circle with a circumference of 40cm = 400/π.

However, I am unable to prove this formula algebraically. I tried using l'hospital's rule after realising that the limit of the original function was 0/0, but I got nowhere. In fact, the result I got was -400π/0, which was very disappointing after so much working out!

I was wondering if anyone could help me prove this algebraically or otherwise. I am happy that I found an equation that proves what I wanted to, but I am unable to prove Wolfram Alpha's result, which is frustrating.

Again, I am new to this forum, so please let me know if I have made any mistakes so that I can edit my question.

calculus limits limits-without-lhopital

calculus limits limits-without-lhopital

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

edited 1 hour ago

Bernard

113k636104

edited 1 hour ago

Bernard

113k636104

edited 1 hour ago

Bernard

113k636104

113k636104

asked 1 hour ago

Eric Spies

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Eric Spies

183

asked 1 hour ago

Eric Spies

183

183

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Eric Spies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
  – jacob1729
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
  – Eric Spies
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
  – jacob1729
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
  – Eric Spies
  1 hour ago
  

  
  
  
  

l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
– jacob1729
1 hour ago

l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression?
– jacob1729
1 hour ago

Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
– Eric Spies
1 hour ago

Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work!
– Eric Spies
1 hour ago

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote



accepted

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

add a comment | 

up vote
1
down vote

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

add a comment | 

up vote
1
down vote

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?

share|cite|improve this answer

answered 1 hour ago

mwt

823115

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Eric Spies is a new contributor. Be nice, and check out our Code of Conduct.

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2941368%2fhow-to-prove-that-the-limit-of-this-equation-is-400-pi%23new-answer', 'question_page');

);

Post as a guest

Name Email

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

add a comment | 

up vote
2
down vote



accepted

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Bernard

113k636104

answered 1 hour ago

Bernard

113k636104

answered 1 hour ago

Bernard

113k636104

113k636104

add a comment | 

add a comment | 

up vote
1
down vote

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

add a comment | 

up vote
1
down vote

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Doug M

40.4k31751

answered 1 hour ago

Doug M

40.4k31751

answered 1 hour ago

Doug M

40.4k31751

40.4k31751

add a comment | 

add a comment | 

up vote
1
down vote

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?

share|cite|improve this answer

answered 1 hour ago

mwt

823115

add a comment | 

up vote
1
down vote

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?

share|cite|improve this answer

answered 1 hour ago

mwt

823115

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?

share|cite|improve this answer

answered 1 hour ago

mwt

823115

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?

share|cite|improve this answer

answered 1 hour ago

mwt

823115

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

mwt

823115

answered 1 hour ago

mwt

823115

answered 1 hour ago

mwt

823115

823115

add a comment | 

add a comment | 

Eric Spies is a new contributor. Be nice, and check out our Code of Conduct.

 

draft saved

draft discarded

Eric Spies is a new contributor. Be nice, and check out our Code of Conduct.

Eric Spies is a new contributor. Be nice, and check out our Code of Conduct.

Eric Spies is a new contributor. Be nice, and check out our Code of Conduct.

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2941368%2fhow-to-prove-that-the-limit-of-this-equation-is-400-pi%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email