Evaluate $lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$Evaluate...
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Evaluate $lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$
Evaluate $lim_{n to infty }frac{(n!)^{1/n}}{n}$.How to evaluate the limit $lim_{x to infty} frac{2^x+1}{2^{x+1}}$Evaluate the limit: $lim_{xto infty} frac{(2x^2 +1)^2}{(x-1)^2(x^2+x)}$Help Evaluating $lim_{xtoinfty}left(cosfrac{3}{x}right)^{x^2}$Evaluate $lim_{ntoinfty} frac{1}{n} left( (n+1)cdots (n+n) right)^frac{1}{n}$Compute $ lim_{xtoinfty}frac{1}{x}int_1^xcosfrac{1}{t},dt $How to find the limit:$lim_{nto infty}left(sum_{k=n+1}^{2n}left(2(2k)^{frac{1}{2k}}-k^{frac{1}{k}}right)-nright)$Evaluate $int_1^{infty} frac {({x}-frac 12)dx}{x}$How to evaluate $ lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n $ where $i=sqrt{-1}$?Evaluate the limit $lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)$
$begingroup$
Calculate and evaluate the limit:
$$lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$$
When plotting the upper and the lower part of the fraction separately it becomes clear that it is a $frac{infty}{infty}$ case. However, I can't solve for the integral. Also, it is not totally clear to me why the limit on the lower part approaches ${infty}$ and not ${0}$ (considering it approaches ${{{infty}^2}{ln(1)}}$). Thank you a lot to everyone that can helps me with it somehow.
calculus limits analysis definite-integrals infinity
edited Mar 18 at 17:52
StubbornAtom
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
$endgroup$
add a comment |
$begingroup$
Calculate and evaluate the limit:
$$lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$$
When plotting the upper and the lower part of the fraction separately it becomes clear that it is a $frac{infty}{infty}$ case. However, I can't solve for the integral. Also, it is not totally clear to me why the limit on the lower part approaches ${infty}$ and not ${0}$ (considering it approaches ${{{infty}^2}{ln(1)}}$). Thank you a lot to everyone that can helps me with it somehow.
calculus limits analysis definite-integrals infinity
edited Mar 18 at 17:52
StubbornAtom
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
$endgroup$
$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16
add a comment |
$begingroup$
Calculate and evaluate the limit:
$$lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$$
When plotting the upper and the lower part of the fraction separately it becomes clear that it is a $frac{infty}{infty}$ case. However, I can't solve for the integral. Also, it is not totally clear to me why the limit on the lower part approaches ${infty}$ and not ${0}$ (considering it approaches ${{{infty}^2}{ln(1)}}$). Thank you a lot to everyone that can helps me with it somehow.
calculus limits analysis definite-integrals infinity
edited Mar 18 at 17:52
StubbornAtom
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
$endgroup$
Calculate and evaluate the limit:
$$lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2lnleft(1+frac{1}{x}right)}$$
When plotting the upper and the lower part of the fraction separately it becomes clear that it is a $frac{infty}{infty}$ case. However, I can't solve for the integral. Also, it is not totally clear to me why the limit on the lower part approaches ${infty}$ and not ${0}$ (considering it approaches ${{{infty}^2}{ln(1)}}$). Thank you a lot to everyone that can helps me with it somehow.
calculus limits analysis definite-integrals infinity
calculus limits analysis definite-integrals infinity
edited Mar 18 at 17:52
StubbornAtom
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
edited Mar 18 at 17:52
StubbornAtom
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
edited Mar 18 at 17:52
StubbornAtom
6,29831440
edited Mar 18 at 17:52
StubbornAtom
6,29831440
edited Mar 18 at 17:52
StubbornAtom
6,29831440
6,29831440
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
asked Mar 17 at 21:06
Lucas CamposLucas Campos
114
114
$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16
add a comment |
$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16
$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Why lower part approaches zero:
$$x^2ln(1+1/x)=frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show:
$$lim_{xto infty}frac{ln(1+1/x)}{1/x^2}=lim_{xto infty}frac{x}{1+1/x}$$
which goes to infinity.
Now, let $$L=lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2ln(1+1/x)}$$
Again, L is of the form $infty/infty$. Again, use L'Hospital's Rule:
$$L=frac{x^2(e^{1/x}-1)-x}{2xtext{ln}(1+1/x)-frac{1}{1+1/x}}$$
Now as mentioned in one of the comments, you can use various expansions to obtain:
$$L=lim_{xto infty}frac{1/2+1/3x+cdots}{1-1/3x^2+cdots}$$
$$L=frac{1}{2}$$
I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Why lower part approaches zero:
$$x^2ln(1+1/x)=frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show:
$$lim_{xto infty}frac{ln(1+1/x)}{1/x^2}=lim_{xto infty}frac{x}{1+1/x}$$
which goes to infinity.
Now, let $$L=lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2ln(1+1/x)}$$
Again, L is of the form $infty/infty$. Again, use L'Hospital's Rule:
$$L=frac{x^2(e^{1/x}-1)-x}{2xtext{ln}(1+1/x)-frac{1}{1+1/x}}$$
Now as mentioned in one of the comments, you can use various expansions to obtain:
$$L=lim_{xto infty}frac{1/2+1/3x+cdots}{1-1/3x^2+cdots}$$
$$L=frac{1}{2}$$
I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Why lower part approaches zero:
$$x^2ln(1+1/x)=frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show:
$$lim_{xto infty}frac{ln(1+1/x)}{1/x^2}=lim_{xto infty}frac{x}{1+1/x}$$
which goes to infinity.
Now, let $$L=lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2ln(1+1/x)}$$
Again, L is of the form $infty/infty$. Again, use L'Hospital's Rule:
$$L=frac{x^2(e^{1/x}-1)-x}{2xtext{ln}(1+1/x)-frac{1}{1+1/x}}$$
Now as mentioned in one of the comments, you can use various expansions to obtain:
$$L=lim_{xto infty}frac{1/2+1/3x+cdots}{1-1/3x^2+cdots}$$
$$L=frac{1}{2}$$
I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Why lower part approaches zero:
$$x^2ln(1+1/x)=frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show:
$$lim_{xto infty}frac{ln(1+1/x)}{1/x^2}=lim_{xto infty}frac{x}{1+1/x}$$
which goes to infinity.
Now, let $$L=lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2ln(1+1/x)}$$
Again, L is of the form $infty/infty$. Again, use L'Hospital's Rule:
$$L=frac{x^2(e^{1/x}-1)-x}{2xtext{ln}(1+1/x)-frac{1}{1+1/x}}$$
Now as mentioned in one of the comments, you can use various expansions to obtain:
$$L=lim_{xto infty}frac{1/2+1/3x+cdots}{1-1/3x^2+cdots}$$
$$L=frac{1}{2}$$
I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
$endgroup$
Why lower part approaches zero:
$$x^2ln(1+1/x)=frac{ln(1+1/x)}{1/x^2}$$which is of the form $0/0$. You can use L'Hospital's Rule to show:
$$lim_{xto infty}frac{ln(1+1/x)}{1/x^2}=lim_{xto infty}frac{x}{1+1/x}$$
which goes to infinity.
Now, let $$L=lim_{xto {infty}} frac{int_1^x (t^2(e^{1/t}-1)-t),dt}{x^2ln(1+1/x)}$$
Again, L is of the form $infty/infty$. Again, use L'Hospital's Rule:
$$L=frac{x^2(e^{1/x}-1)-x}{2xtext{ln}(1+1/x)-frac{1}{1+1/x}}$$
Now as mentioned in one of the comments, you can use various expansions to obtain:
$$L=lim_{xto infty}frac{1/2+1/3x+cdots}{1-1/3x^2+cdots}$$
$$L=frac{1}{2}$$
I might have done some calculation mistake in the very last part, but I hope you know how to handle such problems now...
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
edited Mar 19 at 19:18
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
answered Mar 18 at 19:04
Ankit KumarAnkit Kumar
1,542221
1,542221
add a comment |
add a comment |
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$begingroup$
My professor used to say "Taylor is your friend"
$endgroup$
– Giuseppe Bargagnati
Mar 17 at 23:35
$begingroup$
With CAS answer is: $frac{1}{2}$
$endgroup$
– Mariusz Iwaniuk
Mar 18 at 17:14
$begingroup$
Leibnitz rule maybe☺️
$endgroup$
– Aditya Garg
Mar 18 at 19:16