Mixing
Find lim of $frac{sqrt[n]{n!}}{sqrt{n^3+2n}}$
$begingroup$
Find lim of $frac{sqrt[n]{n!}}{sqrt{n^3+2n}}$
Please help me .Thanks so much
limits
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
$endgroup$
add a comment |
$begingroup$
Find lim of $frac{sqrt[n]{n!}}{sqrt{n^3+2n}}$
Please help me .Thanks so much
limits
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
$endgroup$
1
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
1
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34
add a comment |
$begingroup$
Find lim of $frac{sqrt[n]{n!}}{sqrt{n^3+2n}}$
Please help me .Thanks so much
limits
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
$endgroup$
Find lim of $frac{sqrt[n]{n!}}{sqrt{n^3+2n}}$
Please help me .Thanks so much
limits
limits
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
asked Jan 27 at 9:30
Trương Văn HàoTrương Văn Hào
51
51
1
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
1
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34
add a comment |
1
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
1
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34
1
1
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
1
1
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Clearly, ${n!}le{n^n}$.
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
notice that
$$ 0 < frac{ n!^{1/n} }{ sqrt{n^3+2n} } leq frac{n}{sqrt{n^3+2n}} to0$$
Use the Squeeze rule now
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
add a comment |
$begingroup$
The numerator is about $n/e$
and the denominator is about $n^{3/2}$
so the ratio is about
$dfrac1{en^{1/2}}$
which goes to zero.
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Clearly, ${n!}le{n^n}$.
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
Hint: Clearly, ${n!}le{n^n}$.
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
Hint: Clearly, ${n!}le{n^n}$.
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
Hint: Clearly, ${n!}le{n^n}$.
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 9:34
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
add a comment |
add a comment |
$begingroup$
notice that
$$ 0 < frac{ n!^{1/n} }{ sqrt{n^3+2n} } leq frac{n}{sqrt{n^3+2n}} to0$$
Use the Squeeze rule now
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
add a comment |
$begingroup$
notice that
$$ 0 < frac{ n!^{1/n} }{ sqrt{n^3+2n} } leq frac{n}{sqrt{n^3+2n}} to0$$
Use the Squeeze rule now
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
add a comment |
$begingroup$
notice that
$$ 0 < frac{ n!^{1/n} }{ sqrt{n^3+2n} } leq frac{n}{sqrt{n^3+2n}} to0$$
Use the Squeeze rule now
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
notice that
$$ 0 < frac{ n!^{1/n} }{ sqrt{n^3+2n} } leq frac{n}{sqrt{n^3+2n}} to0$$
Use the Squeeze rule now
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
answered Jan 27 at 9:40
Jimmy SabaterJimmy Sabater
2,539325
2,539325
add a comment |
add a comment |
$begingroup$
The numerator is about $n/e$
and the denominator is about $n^{3/2}$
so the ratio is about
$dfrac1{en^{1/2}}$
which goes to zero.
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
$endgroup$
add a comment |
$begingroup$
The numerator is about $n/e$
and the denominator is about $n^{3/2}$
so the ratio is about
$dfrac1{en^{1/2}}$
which goes to zero.
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
$endgroup$
add a comment |
$begingroup$
The numerator is about $n/e$
and the denominator is about $n^{3/2}$
so the ratio is about
$dfrac1{en^{1/2}}$
which goes to zero.
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
$endgroup$
The numerator is about $n/e$
and the denominator is about $n^{3/2}$
so the ratio is about
$dfrac1{en^{1/2}}$
which goes to zero.
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
answered Jan 27 at 9:48
marty cohenmarty cohen
74.6k549129
74.6k549129
add a comment |
add a comment |
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1
$begingroup$
$n$ tends to infinity here?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 9:34
1
$begingroup$
Is it $nto 0$ ?
$endgroup$
– Claude Leibovici
Jan 27 at 9:34