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Is $displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x = e^5$?

-1

$begingroup$

I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?

limits

share|cite|improve this question

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It will be nice if you show your solution
  $endgroup$
  – haqnatural
  Oct 22 '17 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @haqnatural
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this is the right answer, how do you proved it?
  $endgroup$
  – Dr. Sonnhard Graubner
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got the answer from an online calc
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  expressed the function using e as the base
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  

add a comment | 

-1

$begingroup$

I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?

limits

share|cite|improve this question

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It will be nice if you show your solution
  $endgroup$
  – haqnatural
  Oct 22 '17 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @haqnatural
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this is the right answer, how do you proved it?
  $endgroup$
  – Dr. Sonnhard Graubner
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got the answer from an online calc
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  expressed the function using e as the base
  $endgroup$
  – Paras Khosla
  Oct 22 '17 at 16:10
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to evaluate
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)^x$.
I'm getting the answer $e^5$ using an online caculator. Can anyone suggest a formal solution?

limits

share|cite|improve this question

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

$endgroup$

share|cite|improve this question

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

share|cite|improve this question

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

edited Oct 22 '17 at 16:19

venrey

16011

edited Oct 22 '17 at 16:19

venrey

16011

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

asked Oct 22 '17 at 16:08

Paras KhoslaParas Khosla

3,257627

2 Answers
2

active

oldest

votes

3

$begingroup$

Hint:

$$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$

$$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$

$$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$

Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$

share|cite|improve this answer

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

1

$begingroup$

Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.

Observe
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
And thus it is in $(1)^{infty}$ form
Hence for the limit of this form answer is given by
$$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.

Rest is just calculation.

share|cite|improve this answer

edited Mar 22 at 20:37

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

$endgroup$

add a comment | 

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3

$begingroup$

Hint:

$$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$

$$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$

$$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$

Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$

share|cite|improve this answer

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

3

$begingroup$

Hint:

$$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$

$$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$

$$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$

Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$

share|cite|improve this answer

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

$begingroup$

Hint:

$$left(left(frac{left(x^2+2x+1right)}{left(x^2-3x+2right)}right)^xright)$$

$$=dfrac{((1+x)^x)^2}{(x-1)^x(x-2)^x}$$

$$=dfrac{left(left(1+dfrac1xright)^xright)^2}{left(1-dfrac1xright)^xleft(1-dfrac2xright)^x}$$

Now use $$lim_{ntoinfty}left(1+dfrac1nright)^n=e$$

share|cite|improve this answer

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

answered Oct 22 '17 at 16:12

lab bhattacharjeelab bhattacharjee

228k15159279

1

$begingroup$

Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.

Observe
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
And thus it is in $(1)^{infty}$ form
Hence for the limit of this form answer is given by
$$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.

Rest is just calculation.

share|cite|improve this answer

edited Mar 22 at 20:37

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

$endgroup$

add a comment | 

1

$begingroup$

Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.

Observe
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
And thus it is in $(1)^{infty}$ form
Hence for the limit of this form answer is given by
$$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.

Rest is just calculation.

share|cite|improve this answer

edited Mar 22 at 20:37

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

$endgroup$

add a comment | 

$begingroup$

Well i know it has been more than a year but, as the problem is still open , i am writing the general methodology.

Observe
$displaystylelim _{xto infty }left(dfrac{x^2+2x+1}{x^2-3x+2}right)=1$
And thus it is in $(1)^{infty}$ form
Hence for the limit of this form answer is given by
$$e^{lim _{xto infty }(f(x)-1)(g(x))}$$ where $lim _{xto infty } f(x)^{g(x)}$ must tend to the $(1)^{infty}$ form.

Rest is just calculation.

share|cite|improve this answer

edited Mar 22 at 20:37

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

$endgroup$

share|cite|improve this answer

edited Mar 22 at 20:37

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111

answered Mar 22 at 20:30

NewBornMATHNewBornMATH

549111