Calculate $lim_xto infty(x + ln(fracpi2 - arctan(x))$ using L'hopital's rule Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Evaluating $xe^-x/lambdabig|_0^infty$ with and without L'Hopital's Ruleevaluate $lim_x to 0+ fracx-sin x(x sin x)^3/2$How to evaluate $lim_x to inftyleft(1 + frac2xright)^3x$ using L'Hôpital's rule?L'Hopital's Rule, Factorials, and DerivativesCan de l'Hopital's rule be used in the case $pm frac-inftyinfty$?Finding the limit $lim_xrightarrow 0^+fracint_1^+inftyfrace^-xyquad-1y^3dyln(1+x).$application of L'Hopital's rule?Find the limit using l'Hopital's Rule$lim_x to infty e^x - frace^xx+1$ Application of L'Hopital's RuleCalculating limit of ln(arctan(x)) using chain rule
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Calculate $lim_xto infty(x + ln(fracpi2 - arctan(x))$ using L'hopital's rule
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Evaluating $xe^-x/lambdabig|_0^infty$ with and without L'Hopital's Ruleevaluate $lim_x to 0+ fracx-sin x(x sin x)^3/2$How to evaluate $lim_x to inftyleft(1 + frac2xright)^3x$ using L'Hôpital's rule?L'Hopital's Rule, Factorials, and DerivativesCan de l'Hopital's rule be used in the case $pm frac-inftyinfty$?Finding the limit $lim_xrightarrow 0^+fracint_1^+inftyfrace^-xyquad-1y^3dyln(1+x).$application of L'Hopital's rule?Find the limit using l'Hopital's Rule$lim_x to infty e^x - frace^xx+1$ Application of L'Hopital's RuleCalculating limit of ln(arctan(x)) using chain rule
$begingroup$
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
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
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
edited Apr 1 at 13:39
YuiTo Cheng
2,52341037
asked Apr 1 at 11:49
HellowhatsupHellowhatsup
445
$endgroup$
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
derivatives
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
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
edited Apr 1 at 13:39
YuiTo Cheng
2,52341037
2,52341037
asked Apr 1 at 11:49
HellowhatsupHellowhatsup
445
asked Apr 1 at 11:49
HellowhatsupHellowhatsup
445
asked Apr 1 at 11:49
HellowhatsupHellowhatsup
445
445
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
answered Apr 1 at 11:56
Gabriele CasseseGabriele Cassese
1,226316
$endgroup$
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*
answered Apr 1 at 12:45
trancelocationtrancelocation
14.3k1929
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
answered Apr 1 at 11:56
Gabriele CasseseGabriele Cassese
1,226316
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
answered Apr 1 at 11:56
Gabriele CasseseGabriele Cassese
1,226316
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
answered Apr 1 at 11:56
Gabriele CasseseGabriele Cassese
1,226316
$endgroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
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
answered Apr 1 at 11:56
Gabriele CasseseGabriele Cassese
1,226316
1,226316
add a comment |
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*
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*
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*
answered Apr 1 at 12:45
trancelocationtrancelocation
14.3k1929
$endgroup$
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*
answered Apr 1 at 12:45
trancelocationtrancelocation
14.3k1929
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
14.3k1929
add a comment |
add a comment |
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