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Calculate $lim_xto infty(x + ln(fracpi2 - arctan(x))$ using L'hopital's rule

1

$begingroup$

I'm new to L'hopital's rule. I know i need to convert it to $fracinftyinfty$ or $frac00$. But I have no idea how to convert the following equation. Thanks in advance for your help!

$$lim_xto infty(x + ln(fracpi2 - arctan(x))$$

derivatives

share|cite|improve this question

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

$endgroup$

add a comment | 

1

$begingroup$

I'm new to L'hopital's rule. I know i need to convert it to $fracinftyinfty$ or $frac00$. But I have no idea how to convert the following equation. Thanks in advance for your help!

$$lim_xto infty(x + ln(fracpi2 - arctan(x))$$

derivatives

share|cite|improve this question

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

$endgroup$

add a comment | 

$begingroup$

I'm new to L'hopital's rule. I know i need to convert it to $fracinftyinfty$ or $frac00$. But I have no idea how to convert the following equation. Thanks in advance for your help!

$$lim_xto infty(x + ln(fracpi2 - arctan(x))$$

derivatives

share|cite|improve this question

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

$endgroup$

share|cite|improve this question

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

share|cite|improve this question

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

edited Apr 1 at 13:39

YuiTo Cheng

2,52341037

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

asked Apr 1 at 11:49

HellowhatsupHellowhatsup

445

2 Answers
2

active

oldest

votes

4

$begingroup$

Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$

share|cite|improve this answer

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

$endgroup$

add a comment | 

0

$begingroup$

You may also proceed as follows using

• $operatornamearccotx = fracpi2 - arctan x$


• 
  $x = cot u$ while considering the limit for $u to 0^+$
  


• $sin u ln u = fracsin uucdot underbraceu ln u_=fracln ufrac1ustackrelL'Hosp.sim-u stackrelu to 0^+longrightarrow 0$

begineqnarray* lim_xto infty(x + ln(fracpi2 - arctan(x))
& stackrelx=cot u= & cot u + ln u \
& = & fraccolorblueoverbracecos u + sin u ln u^stackrelu to 0^+longrightarrow1sin u\
& stackrelu to 0^+longrightarrow & +infty
endeqnarray*

share|cite|improve this answer

edited Apr 1 at 13:09

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

$endgroup$

add a comment | 

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4

$begingroup$

Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$

share|cite|improve this answer

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

$endgroup$

add a comment | 

4

$begingroup$

Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$

share|cite|improve this answer

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

$endgroup$

add a comment | 

$begingroup$

Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$

share|cite|improve this answer

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

$endgroup$

share|cite|improve this answer

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

answered Apr 1 at 11:56

Gabriele CasseseGabriele Cassese

1,226316

0

$begingroup$

You may also proceed as follows using

• $operatornamearccotx = fracpi2 - arctan x$


• 
  $x = cot u$ while considering the limit for $u to 0^+$
  


• $sin u ln u = fracsin uucdot underbraceu ln u_=fracln ufrac1ustackrelL'Hosp.sim-u stackrelu to 0^+longrightarrow 0$

begineqnarray* lim_xto infty(x + ln(fracpi2 - arctan(x))
& stackrelx=cot u= & cot u + ln u \
& = & fraccolorblueoverbracecos u + sin u ln u^stackrelu to 0^+longrightarrow1sin u\
& stackrelu to 0^+longrightarrow & +infty
endeqnarray*

share|cite|improve this answer

edited Apr 1 at 13:09

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

$endgroup$

add a comment | 

0

$begingroup$

You may also proceed as follows using

• $operatornamearccotx = fracpi2 - arctan x$


• 
  $x = cot u$ while considering the limit for $u to 0^+$
  


• $sin u ln u = fracsin uucdot underbraceu ln u_=fracln ufrac1ustackrelL'Hosp.sim-u stackrelu to 0^+longrightarrow 0$

begineqnarray* lim_xto infty(x + ln(fracpi2 - arctan(x))
& stackrelx=cot u= & cot u + ln u \
& = & fraccolorblueoverbracecos u + sin u ln u^stackrelu to 0^+longrightarrow1sin u\
& stackrelu to 0^+longrightarrow & +infty
endeqnarray*

share|cite|improve this answer

edited Apr 1 at 13:09

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

$endgroup$

add a comment | 

$begingroup$

You may also proceed as follows using

• $operatornamearccotx = fracpi2 - arctan x$


• 
  $x = cot u$ while considering the limit for $u to 0^+$
  


• $sin u ln u = fracsin uucdot underbraceu ln u_=fracln ufrac1ustackrelL'Hosp.sim-u stackrelu to 0^+longrightarrow 0$

begineqnarray* lim_xto infty(x + ln(fracpi2 - arctan(x))
& stackrelx=cot u= & cot u + ln u \
& = & fraccolorblueoverbracecos u + sin u ln u^stackrelu to 0^+longrightarrow1sin u\
& stackrelu to 0^+longrightarrow & +infty
endeqnarray*

share|cite|improve this answer

edited Apr 1 at 13:09

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

$endgroup$

share|cite|improve this answer

edited Apr 1 at 13:09

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929

answered Apr 1 at 12:45

trancelocationtrancelocation

14.3k1929