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Mathematics 14 Online
OpenStudy (anonymous):

whats the hospital rule for ln 3x

OpenStudy (anonymous):

is it 1/3?

OpenStudy (anonymous):

drink plenty of fluids, stay out of the sun

OpenStudy (anonymous):

lol thanks i havent been out too extremely hot

OpenStudy (anonymous):

ahaha. Do you men l'Hospitals rule? Is this supposed to be the limit as x->infinity of ln(3/x)

OpenStudy (anonymous):

i have this here: lim x→ ln 3x 3x

OpenStudy (anonymous):

x approaches infinity ln3x/\[\sqrt{3x}\]

OpenStudy (anonymous):

so i dont know how i would compute that

OpenStudy (anonymous):

\[\lim_{x\rightarrow\infty}\frac{\ln(3x)}{\sqrt{3x}}\]

OpenStudy (anonymous):

is that it?

OpenStudy (anonymous):

if so use no rule. get 0 from your eyeballs. log grows very slow, much slower than \[\sqrt{x}\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

you are of course welcome to use l'hopital if you like

OpenStudy (anonymous):

get \[\frac{1}{x}\times \frac{2\sqrt{x}}{3}\] and then get \[\frac{2}{3\sqrt{x}}\] take the limit, get 0

OpenStudy (anonymous):

ok so thats the final answers?

OpenStudy (anonymous):

clear yes? derivative of \[\ln(3x)\] is \[\frac{1}{x}\] and derivative of \[\sqrt{3x}\] is \[\frac{3}{2\sqrt{x}}\]

OpenStudy (anonymous):

yes answer is 0

OpenStudy (anonymous):

yes thanks!

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