Fdtxjr

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Could someone tell me why $log n 0$? [on hold]

Can someone explain log?How can I calculate $lim_x to 0 log(cos(x))/log(cos(3x))$ without l'Hopital?Proving limit with $log(n!)$Simplyfing Log QuestionHow to compute $=lim_n to infty Big( fraclog(n+1)log(n) cdot fracn-2n-1 Big)$“by hand”?Why does $(log n)^log n = Omega(n^10)$?Why do $2log x$ and $log x^2$ look different when graphing?Value of limit $(1+e^-x)^2^x logx$Find $lim_nrightarrowinfty fraclog(2 + 3^n)2n$Solving $2log(n!) - nlog n le 2nlog n$

-3

$begingroup$

To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
$$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$

Could some one tell me how to convert the expression 1 to 2?
I don't understand.
Thank you!

logarithms

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edited Mar 11 at 13:34

Rócherz

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put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03

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-3

$begingroup$

To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
$$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$

Could some one tell me how to convert the expression 1 to 2?
I don't understand.
Thank you!

logarithms

share|cite|improve this question

edited Mar 11 at 13:34

Rócherz

2,9512821

asked Mar 11 at 10:18

WU NICKWU NICK

1

New contributor

WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會 Mar 11 at 21:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Gibbs, José Carlos Santos, ancientmathematician, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

To show that $lim_ntoinftyfracn^alog n = infty$ you can apply L'Hôpital.
$$lim_ntoinfty fracn^alog n = lim_ntoinfty fracan^a-1frac 1n = lim_ntoinfty acdot n^a = infty, (because a>0).$$

Could some one tell me how to convert the expression 1 to 2?
I don't understand.
Thank you!

logarithms

share|cite|improve this question

edited Mar 11 at 13:34

Rócherz

2,9512821

asked Mar 11 at 10:18

WU NICKWU NICK

1

New contributor

WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 11 at 13:34

Rócherz

2,9512821

asked Mar 11 at 10:18

WU NICKWU NICK

1

New contributor

WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 11 at 13:34

Rócherz

2,9512821

asked Mar 11 at 10:18

WU NICKWU NICK

1

New contributor

WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 11 at 13:34

Rócherz

2,9512821

edited Mar 11 at 13:34

Rócherz

2,9512821

asked Mar 11 at 10:18

WU NICKWU NICK

1

New contributor

WU NICK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 10:18

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1

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If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$

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answered Mar 11 at 10:25

FredFred

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1

$begingroup$

If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$

share|cite|improve this answer

answered Mar 11 at 10:25

FredFred

48.3k1849

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add a comment | 

1

$begingroup$

If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$

share|cite|improve this answer

answered Mar 11 at 10:25

FredFred

48.3k1849

$endgroup$

add a comment | 

$begingroup$

If $f(x)=x^a$ and $g(x)= log x$, then $ fracf'(x)g'(x)= fracax^a-1frac1x=ax^a.$

share|cite|improve this answer

answered Mar 11 at 10:25

FredFred

48.3k1849

$endgroup$

share|cite|improve this answer

answered Mar 11 at 10:25

FredFred

48.3k1849

answered Mar 11 at 10:25

FredFred

48.3k1849

answered Mar 11 at 10:25

FredFred

48.3k1849