Limit to compare growth of function
$begingroup$
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 Nov 15 '18 at 9:51
user376343
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
$endgroup$
add a comment |
$begingroup$
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 Nov 15 '18 at 9:51
user376343
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
$endgroup$
add a comment |
$begingroup$
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 Nov 15 '18 at 9:51
user376343
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
$endgroup$
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 Nov 15 '18 at 9:51
user376343
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
edited Nov 15 '18 at 9:51
user376343
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
edited Nov 15 '18 at 9:51
user376343
3,8483829
edited Nov 15 '18 at 9:51
user376343
3,8483829
edited Nov 15 '18 at 9:51
user376343
3,8483829
3,8483829
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
asked Nov 15 '18 at 5:36
user3767495user3767495
4078
4078
add a comment |
add a comment |
2 Answers
2
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$begingroup$
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 Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
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add a comment |
$begingroup$
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
$endgroup$
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
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 Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
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 Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
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 Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
answered Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
answered Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
answered Nov 15 '18 at 6:06
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
add a comment |
add a comment |
$begingroup$
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
$endgroup$
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
add a comment |
$begingroup$
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
$endgroup$
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
add a comment |
$begingroup$
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
$endgroup$
$(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
answered Nov 15 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
answered Nov 15 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
answered Nov 15 '18 at 6:00
Kavi Rama MurthyKavi Rama Murthy
62.5k42262
62.5k42262
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
add a comment |
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
$begingroup$
Same time, same answer !
$endgroup$
– Claude Leibovici
Nov 15 '18 at 6:06
add a comment |
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