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anonymous
 3 years ago
determine if existent the limit:
lim >0 sinxx/x^3
anonymous
 3 years ago
determine if existent the limit: lim >0 sinxx/x^3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you put parenthesis to make the problem a little more clear?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it \[\lim_{x \rightarrow 0} \frac{ sinxx }{ x^{3} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes that is the exact question.. @Eda2012

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you use :'hopital's rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, we can use Hopital rule!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that makes it so much easier @Dido525 thanks for thinking of that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay then! Take the derivative of the top and bottom terms :) . But before that check if the limit is 0/0 or infinity/ infinity.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haha. I always ask ^_^ .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah the limit is 0/0 if you directly sub in.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I need to show step by step to the professor :( lol... can anybody do that for me.. sorry guys

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352266897769:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[1^{st} \] \[\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx1 }{ 3x ^{2} }\] \[2^{nd}\] \[\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ sinx }{ 6x }\] \[3^{rd}\] \[\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ cosx }{ 6 }\] \[=\frac{ 1 }{ 6 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah. You often have to use L'hopital's rule multiple times.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thnx all... what will be the solution if we use the rule L'hopital? since that is the required

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the L'hopital's rule ...u must not get \[\frac{ \infty }{ \infty } or \frac{ 0 }{ 0 }\] when u plug in the lim of x....because the equation will become undefined for every value of x...that's why u have to use the rule many times until u get the actual value that can be defined....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352268277086:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352268373043:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 ...r u using the divergent test...???
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