$ limlimits_xrightarrow +infty left(frac2pi arctan x right)^x$ and $lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)$?Compute $lim limits_xtoinfty (fracx-2x+2)^x$L'Hopital's Rule with $lim limits_x to inftyfrac2^xe^left(x^2right)$How to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Evaluate $limlimits_xtoinftyx(fracpi2-arctan(x))$ without using L'HôpitalCalculate this limit : $lim_xrightarrow +inftyleft[xleft(4arctanleft(fracx+1xright)-piright)right]$Prove that $limlimits_nrightarrowinfty left(1+frac1a_n right)^a_n=e$ if $limlimits_nrightarrowinfty a_n=infty$Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$How would you calculate this limit? $limlimits_n rightarrowinftyfracpi2nsumlimits_k=1^ncosleft(fracpi2nkright)$Why am I computing $lim limits_x to infty x left(arctan fracx+1x+2 -arctan fracxx+2 right)$ wrong?
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$ limlimits_xrightarrow +infty left(frac2pi arctan x right)^x$ and $lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)$?
Compute $lim limits_xtoinfty (fracx-2x+2)^x$L'Hopital's Rule with $lim limits_x to inftyfrac2^xe^left(x^2right)$How to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Evaluate $limlimits_xtoinftyx(fracpi2-arctan(x))$ without using L'HôpitalCalculate this limit : $lim_xrightarrow +inftyleft[xleft(4arctanleft(fracx+1xright)-piright)right]$Prove that $limlimits_nrightarrowinfty left(1+frac1a_n right)^a_n=e$ if $limlimits_nrightarrowinfty a_n=infty$Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$How would you calculate this limit? $limlimits_n rightarrowinftyfracpi2nsumlimits_k=1^ncosleft(fracpi2nkright)$Why am I computing $lim limits_x to infty x left(arctan fracx+1x+2 -arctan fracxx+2 right)$ wrong?
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
calculus limits
edited 2 days ago
user21820
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
$endgroup$
add a comment |
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
calculus limits
edited 2 days ago
user21820
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
$endgroup$
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
2
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
$begingroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
calculus limits
edited 2 days ago
user21820
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
$endgroup$
I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$
For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.
For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?
calculus limits
calculus limits
edited 2 days ago
user21820
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
edited 2 days ago
user21820
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
edited 2 days ago
user21820
40k544160
edited 2 days ago
user21820
40k544160
edited 2 days ago
user21820
40k544160
40k544160
asked 2 days ago
lanse7ptylanse7pty
1,8461823
asked 2 days ago
lanse7ptylanse7pty
1,8461823
asked 2 days ago
lanse7ptylanse7pty
1,8461823
1,8461823
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
2
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
2
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
2
2
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 2 days ago
trancelocationtrancelocation
13.5k1828
$endgroup$
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
For the first: taking $log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
lim_xto+inftyxlogleft(frac2piarctan x right) =
lim_tto 0^+frac 1tlogleft(frac2piarctan(1/t)right) =
lim_tto 0^+frac -1(t^2 + 1)arctan(1/t)) = -frac 2pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 2 days ago
trancelocationtrancelocation
13.5k1828
$endgroup$
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 2 days ago
trancelocationtrancelocation
13.5k1828
$endgroup$
add a comment |
$begingroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 2 days ago
trancelocationtrancelocation
13.5k1828
$endgroup$
You can solve the first one using
- $arctan x + operatornamearccotx = fracpi2$
- $lim_yto 0(1-y)^1/y = e^-1$
- $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$
begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*
The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider
$fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.
answered 2 days ago
trancelocationtrancelocation
13.5k1828
answered 2 days ago
trancelocationtrancelocation
13.5k1828
answered 2 days ago
trancelocationtrancelocation
13.5k1828
answered 2 days ago
trancelocationtrancelocation
13.5k1828
13.5k1828
add a comment |
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
$endgroup$
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
$endgroup$
add a comment |
$begingroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
$endgroup$
Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
edited 2 days ago
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
answered 2 days ago
Paras KhoslaParas Khosla
2,782423
2,782423
add a comment |
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$
Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 2 days ago
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
add a comment |
add a comment |
$begingroup$
For the first: taking $log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
lim_xto+inftyxlogleft(frac2piarctan x right) =
lim_tto 0^+frac 1tlogleft(frac2piarctan(1/t)right) =
lim_tto 0^+frac -1(t^2 + 1)arctan(1/t)) = -frac 2pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
add a comment |
$begingroup$
For the first: taking $log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
lim_xto+inftyxlogleft(frac2piarctan x right) =
lim_tto 0^+frac 1tlogleft(frac2piarctan(1/t)right) =
lim_tto 0^+frac -1(t^2 + 1)arctan(1/t)) = -frac 2pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
add a comment |
$begingroup$
For the first: taking $log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
lim_xto+inftyxlogleft(frac2piarctan x right) =
lim_tto 0^+frac 1tlogleft(frac2piarctan(1/t)right) =
lim_tto 0^+frac -1(t^2 + 1)arctan(1/t)) = -frac 2pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
$endgroup$
For the first: taking $log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
lim_xto+inftyxlogleft(frac2piarctan x right) =
lim_tto 0^+frac 1tlogleft(frac2piarctan(1/t)right) =
lim_tto 0^+frac -1(t^2 + 1)arctan(1/t)) = -frac 2pi.
$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35k42971
35k42971
add a comment |
add a comment |
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$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 days ago
2
$begingroup$
Try to avoid asking several questions in a single post. If you feel that the answer would be similar, ask about the first limit, and then try to solve the second one by yourself, and if you are still stuck, ask a separate question. The software has a limit on how many questions you can ask in a single day, and doing it like this is circumventing these limitations.
$endgroup$
– Asaf Karagila♦
2 days ago