Limit to compare growth of function
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I wanted to compare growth of two functions
$F_1:n^,lg,lg n$
$F_2:(3/2)^n$
$lim_n to infty fracn^lglg n(3/2)^n$
After differentiating it $lg , lg n$ times I get
$lim_n to infty frac(lglg n)!(lg(3/2))^lglg n(3/2)^n$
How do I proceed forward?
limits
edited yesterday
user376343
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asked 2 days ago
user3767495
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up vote
2
down vote
favorite
I wanted to compare growth of two functions
$F_1:n^,lg,lg n$
$F_2:(3/2)^n$
$lim_n to infty fracn^lglg n(3/2)^n$
After differentiating it $lg , lg n$ times I get
$lim_n to infty frac(lglg n)!(lg(3/2))^lglg n(3/2)^n$
How do I proceed forward?
limits
edited yesterday
user376343
2,1441716
asked 2 days ago
user3767495
1348
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I wanted to compare growth of two functions
$F_1:n^,lg,lg n$
$F_2:(3/2)^n$
$lim_n to infty fracn^lglg n(3/2)^n$
After differentiating it $lg , lg n$ times I get
$lim_n to infty frac(lglg n)!(lg(3/2))^lglg n(3/2)^n$
How do I proceed forward?
limits
edited yesterday
user376343
2,1441716
asked 2 days ago
user3767495
1348
I wanted to compare growth of two functions
$F_1:n^,lg,lg n$
$F_2:(3/2)^n$
$lim_n to infty fracn^lglg n(3/2)^n$
After differentiating it $lg , lg n$ times I get
$lim_n to infty frac(lglg n)!(lg(3/2))^lglg n(3/2)^n$
How do I proceed forward?
limits
limits
edited yesterday
user376343
2,1441716
asked 2 days ago
user3767495
1348
edited yesterday
user376343
2,1441716
asked 2 days ago
user3767495
1348
edited yesterday
user376343
2,1441716
edited yesterday
user376343
2,1441716
edited yesterday
user376343
2,1441716
2,1441716
asked 2 days ago
user3767495
1348
asked 2 days ago
user3767495
1348
asked 2 days ago
user3767495
1348
1348
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
3
down vote
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ because $frac (ln ln n) (ln n) n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^(ln ln n) (ln n) /(3/2)^n to 0$. This is same as $frac n^ln ln n (3/2)^n to 0$
answered yesterday
Kavi Rama Murthy
39.3k31748
Same time, same answer !
– Claude Leibovici
yesterday
add a comment |
up vote
3
down vote
Consider
$$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
$$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
$$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$
answered yesterday
Claude Leibovici
116k1156131
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ because $frac (ln ln n) (ln n) n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^(ln ln n) (ln n) /(3/2)^n to 0$. This is same as $frac n^ln ln n (3/2)^n to 0$
answered yesterday
Kavi Rama Murthy
39.3k31748
Same time, same answer !
– Claude Leibovici
yesterday
add a comment |
up vote
3
down vote
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ because $frac (ln ln n) (ln n) n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^(ln ln n) (ln n) /(3/2)^n to 0$. This is same as $frac n^ln ln n (3/2)^n to 0$
answered yesterday
Kavi Rama Murthy
39.3k31748
Same time, same answer !
– Claude Leibovici
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ because $frac (ln ln n) (ln n) n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^(ln ln n) (ln n) /(3/2)^n to 0$. This is same as $frac n^ln ln n (3/2)^n to 0$
answered yesterday
Kavi Rama Murthy
39.3k31748
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ because $frac (ln ln n) (ln n) n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^(ln ln n) (ln n) /(3/2)^n to 0$. This is same as $frac n^ln ln n (3/2)^n to 0$
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
39.3k31748
Same time, same answer !
– Claude Leibovici
yesterday
add a comment |
Same time, same answer !
– Claude Leibovici
yesterday
Same time, same answer !
– Claude Leibovici
yesterday
Same time, same answer !
– Claude Leibovici
yesterday
add a comment |
up vote
3
down vote
Consider
$$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
$$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
$$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$
answered yesterday
Claude Leibovici
116k1156131
add a comment |
up vote
3
down vote
Consider
$$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
$$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
$$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$
answered yesterday
Claude Leibovici
116k1156131
add a comment |
up vote
3
down vote
up vote
3
down vote
Consider
$$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
$$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
$$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$
answered yesterday
Claude Leibovici
116k1156131
Consider
$$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
$$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
$$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$
answered yesterday
Claude Leibovici
116k1156131
answered yesterday
Claude Leibovici
116k1156131
answered yesterday
Claude Leibovici
116k1156131
answered yesterday
Claude Leibovici
116k1156131
116k1156131
add a comment |
add a comment |
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