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OpenStudy (anonymous):
whats the hospital rule for ln 3x
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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
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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
\[\sqrt{x}\]
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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
\[\frac{3}{2\sqrt{x}}\]
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OpenStudy (anonymous):
yes answer is 0
OpenStudy (anonymous):
yes thanks!
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