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etemplinBest ResponseYou've already chosen the best response.0
can you put parenthesis to make the problem a little more clear?
 one year ago

Eda2012Best ResponseYou've already chosen the best response.0
is it \[\lim_{x \rightarrow 0} \frac{ sinxx }{ x^{3} }\]
 one year ago

kabulisBest ResponseYou've already chosen the best response.0
Yes that is the exact question.. @Eda2012
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Can you use :'hopital's rule?
 one year ago

kabulisBest ResponseYou've already chosen the best response.0
Yes, we can use Hopital rule!
 one year ago

etemplinBest ResponseYou've already chosen the best response.0
that makes it so much easier @Dido525 thanks for thinking of that
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Okay then! Take the derivative of the top and bottom terms :) . But before that check if the limit is 0/0 or infinity/ infinity.
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Haha. I always ask ^_^ .
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Yeah the limit is 0/0 if you directly sub in.
 one year ago

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

Eda2012Best ResponseYou've already chosen the best response.0
\[1^{st} \] \[\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx1 }{ 3x ^{2} }\] \[2^{nd}\] \[\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ sinx }{ 6x }\] \[3^{rd}\] \[\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ cosx }{ 6 }\] \[=\frac{ 1 }{ 6 }\]
 one year ago

Dido525Best ResponseYou've already chosen the best response.1
Yeah. You often have to use L'hopital's rule multiple times.
 one year ago

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

Eda2012Best ResponseYou've already chosen the best response.0
the 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....
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1352268277086:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1352268373043:dw
 one year ago

Eda2012Best ResponseYou've already chosen the best response.0
@mahmit2012 ...r u using the divergent test...???
 one year ago
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