Limit to compare growth of function
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I wanted to compare growth of two functions
$F_1:n^{,lg,lg n}$
$F_2:(3/2)^n$
$lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$
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 Nov 15 at 9:51
user376343
2,2531716
asked Nov 15 at 5:36
user3767495
1518
add a comment |
up vote
2
down vote
favorite
I wanted to compare growth of two functions
$F_1:n^{,lg,lg n}$
$F_2:(3/2)^n$
$lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$
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 Nov 15 at 9:51
user376343
2,2531716
asked Nov 15 at 5:36
user3767495
1518
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I wanted to compare growth of two functions
$F_1:n^{,lg,lg n}$
$F_2:(3/2)^n$
$lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$
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 Nov 15 at 9:51
user376343
2,2531716
asked Nov 15 at 5:36
user3767495
1518
I wanted to compare growth of two functions
$F_1:n^{,lg,lg n}$
$F_2:(3/2)^n$
$lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$
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 Nov 15 at 9:51
user376343
2,2531716
asked Nov 15 at 5:36
user3767495
1518
edited Nov 15 at 9:51
user376343
2,2531716
asked Nov 15 at 5:36
user3767495
1518
edited Nov 15 at 9:51
user376343
2,2531716
edited Nov 15 at 9:51
user376343
2,2531716
edited Nov 15 at 9:51
user376343
2,2531716
2,2531716
asked Nov 15 at 5:36
user3767495
1518
asked Nov 15 at 5:36
user3767495
1518
asked Nov 15 at 5:36
user3767495
1518
1518
add a comment |
add a comment |
2 Answers
2
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 Nov 15 at 6:00
Kavi Rama Murthy
41k31751
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
add a comment |
up vote
3
down vote
Consider
$$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
$$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
$$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$
answered Nov 15 at 6:06
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 Nov 15 at 6:00
Kavi Rama Murthy
41k31751
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
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 Nov 15 at 6:00
Kavi Rama Murthy
41k31751
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
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 Nov 15 at 6:00
Kavi Rama Murthy
41k31751
$(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ 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 Nov 15 at 6:00
Kavi Rama Murthy
41k31751
answered Nov 15 at 6:00
Kavi Rama Murthy
41k31751
answered Nov 15 at 6:00
Kavi Rama Murthy
41k31751
answered Nov 15 at 6:00
Kavi Rama Murthy
41k31751
41k31751
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
add a comment |
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
Same time, same answer !
– Claude Leibovici
Nov 15 at 6:06
add a comment |
up vote
3
down vote
Consider
$$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
$$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
$$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
add a comment |
up vote
3
down vote
Consider
$$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
$$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
$$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
add a comment |
up vote
3
down vote
up vote
3
down vote
Consider
$$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
$$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
$$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
Consider
$$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
$$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
$$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
answered Nov 15 at 6:06
Claude Leibovici
116k1156131
116k1156131
add a comment |
add a comment |
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