Mixing
Calculate $lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$
$begingroup$
$$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?
calculus limits
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
$endgroup$
add a comment |
$begingroup$
$$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?
calculus limits
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
$endgroup$
add a comment |
$begingroup$
$$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?
calculus limits
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
$endgroup$
$$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?
calculus limits
calculus limits
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
edited Jan 26 at 19:28
TheSimpliFire
12.8k62461
12.8k62461
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
asked Jan 26 at 18:46
DaniVajaDaniVaja
977
977
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?
answered Jan 26 at 18:53
BernardBernard
123k741117
$endgroup$
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
add a comment |
$begingroup$
Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
$endgroup$
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
add a comment |
$begingroup$
Making the problem more general, consider
$$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
$$logleft(frac{n^{2}+a}{n+b}right)=log
left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
$${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
left({n}right)right)+log ^2left({n}right)}{2
n^2}+Oleft(frac{1}{n^3}right)$$
Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
$begin{array}{l}
displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
lim_{n to infty}n^{1/n} =
expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
displaystyle
expleft(lim_{n to infty}
{lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
expleft(lim_{n to infty}
lnleft(1 + {1 over n}right)right) = expleft(0right) =
bbox[10px,#ffd,border:1px groove navy]{1}
end{array}
$
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?
answered Jan 26 at 18:53
BernardBernard
123k741117
$endgroup$
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
add a comment |
$begingroup$
Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?
answered Jan 26 at 18:53
BernardBernard
123k741117
$endgroup$
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
add a comment |
$begingroup$
Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?
answered Jan 26 at 18:53
BernardBernard
123k741117
$endgroup$
Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?
answered Jan 26 at 18:53
BernardBernard
123k741117
edited Jan 26 at 18:57
answered Jan 26 at 18:53
BernardBernard
123k741117
answered Jan 26 at 18:53
BernardBernard
123k741117
answered Jan 26 at 18:53
BernardBernard
123k741117
123k741117
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
add a comment |
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
1
1
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
$endgroup$
– DaniVaja
Jan 26 at 19:00
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
$begingroup$
Yes, absolutely.
$endgroup$
– Bernard
Jan 26 at 19:02
add a comment |
$begingroup$
Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
$endgroup$
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
add a comment |
$begingroup$
Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
$endgroup$
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
add a comment |
$begingroup$
Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
$endgroup$
Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
answered Jan 26 at 19:00
J.G.J.G.
31.5k23149
31.5k23149
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
add a comment |
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
$begingroup$
This is the same as what I propose, with more details.
$endgroup$
– Bernard
Jan 26 at 19:01
add a comment |
$begingroup$
Making the problem more general, consider
$$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
$$logleft(frac{n^{2}+a}{n+b}right)=log
left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
$${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
left({n}right)right)+log ^2left({n}right)}{2
n^2}+Oleft(frac{1}{n^3}right)$$
Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Making the problem more general, consider
$$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
$$logleft(frac{n^{2}+a}{n+b}right)=log
left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
$${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
left({n}right)right)+log ^2left({n}right)}{2
n^2}+Oleft(frac{1}{n^3}right)$$
Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Making the problem more general, consider
$$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
$$logleft(frac{n^{2}+a}{n+b}right)=log
left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
$${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
left({n}right)right)+log ^2left({n}right)}{2
n^2}+Oleft(frac{1}{n^3}right)$$
Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Making the problem more general, consider
$$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
$$logleft(frac{n^{2}+a}{n+b}right)=log
left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
$${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
left({n}right)right)+log ^2left({n}right)}{2
n^2}+Oleft(frac{1}{n^3}right)$$
Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 3:06
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
add a comment |
add a comment |
$begingroup$
$begin{array}{l}
displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
lim_{n to infty}n^{1/n} =
expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
displaystyle
expleft(lim_{n to infty}
{lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
expleft(lim_{n to infty}
lnleft(1 + {1 over n}right)right) = expleft(0right) =
bbox[10px,#ffd,border:1px groove navy]{1}
end{array}
$
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$begin{array}{l}
displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
lim_{n to infty}n^{1/n} =
expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
displaystyle
expleft(lim_{n to infty}
{lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
expleft(lim_{n to infty}
lnleft(1 + {1 over n}right)right) = expleft(0right) =
bbox[10px,#ffd,border:1px groove navy]{1}
end{array}
$
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$begin{array}{l}
displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
lim_{n to infty}n^{1/n} =
expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
displaystyle
expleft(lim_{n to infty}
{lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
expleft(lim_{n to infty}
lnleft(1 + {1 over n}right)right) = expleft(0right) =
bbox[10px,#ffd,border:1px groove navy]{1}
end{array}
$
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$begin{array}{l}
displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
lim_{n to infty}n^{1/n} =
expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
displaystyle
expleft(lim_{n to infty}
{lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
expleft(lim_{n to infty}
lnleft(1 + {1 over n}right)right) = expleft(0right) =
bbox[10px,#ffd,border:1px groove navy]{1}
end{array}
$
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
edited Jan 28 at 6:42
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
answered Jan 28 at 2:27
Felix MarinFelix Marin
68.8k7109146
68.8k7109146
add a comment |
add a comment |
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