Calculate $lim_xto infty(x + ln(fracpi2 - arctan(x))$ using L'hopital's rule The Next CEO of Stack OverflowEvaluating $xe^-x/lambdabig|_0^infty$ 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
The Next CEO of Stack OverflowEvaluating $xe^-x/lambdabig|_0^infty$ 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 14 hours ago
YuiTo Cheng
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
$endgroup$
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 14 hours ago
YuiTo Cheng
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
$endgroup$
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 14 hours ago
YuiTo Cheng
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
$endgroup$
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 14 hours ago
YuiTo Cheng
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
edited 14 hours ago
YuiTo Cheng
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
edited 14 hours ago
YuiTo Cheng
2,1863937
edited 14 hours ago
YuiTo Cheng
2,1863937
edited 14 hours ago
YuiTo Cheng
2,1863937
2,1863937
asked 16 hours ago
HellowhatsupHellowhatsup
184
asked 16 hours ago
HellowhatsupHellowhatsup
184
asked 16 hours ago
HellowhatsupHellowhatsup
184
184
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 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
$endgroup$
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 15 hours ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
$endgroup$
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 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
$endgroup$
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 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
$endgroup$
Hint:
Consider $x+lnleft(fracpi2-arctan(x)right)=\
-ln(e^-x)+lnleft(fracpi2-arctan(x)right)=\
=lnleft(fracfracpi2-arctan(x)e^-xright)$
answered 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
answered 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
answered 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
answered 16 hours ago
Gabriele CasseseGabriele Cassese
1,211316
1,211316
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 15 hours ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 15 hours ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 15 hours ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 15 hours ago
trancelocationtrancelocation
13.4k1827
edited 15 hours ago
answered 15 hours ago
trancelocationtrancelocation
13.4k1827
answered 15 hours ago
trancelocationtrancelocation
13.4k1827
answered 15 hours ago
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
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