How to calculate the two limits? The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$
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How to calculate the two limits?
The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$
$begingroup$
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 ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
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 ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
$begingroup$
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 hours 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 ?
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
$endgroup$
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 ?
limits
limits
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
edited 1 hour ago
lanse7pty
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
asked 2 hours ago
lanse7ptylanse7pty
1,8361823
1,8361823
$begingroup$
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 hours 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 hours 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 hours 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 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 hours ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
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 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
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 1 min ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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
$begingroup$
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 hours ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
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 hours ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
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 hours ago
Paras KhoslaParas Khosla
2,726423
$endgroup$
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 hours ago
Paras KhoslaParas Khosla
2,726423
edited 1 hour ago
answered 2 hours ago
Paras KhoslaParas Khosla
2,726423
answered 2 hours ago
Paras KhoslaParas Khosla
2,726423
answered 2 hours ago
Paras KhoslaParas Khosla
2,726423
2,726423
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
answered 1 hour ago
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
add a comment |
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
add a comment |
$begingroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
$endgroup$
I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
answered 2 hours ago
AdmuthAdmuth
185
185
add a comment |
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 1 min ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 1 min ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 1 min ago
trancelocationtrancelocation
13.4k1827
$endgroup$
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 1 min ago
trancelocationtrancelocation
13.4k1827
answered 1 min ago
trancelocationtrancelocation
13.4k1827
answered 1 min ago
trancelocationtrancelocation
13.4k1827
answered 1 min ago
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
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Required, but never shown
$begingroup$
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...
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– Arturo Magidin
2 hours ago