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etemplin Group TitleBest ResponseYou've already chosen the best response.0
can you put parenthesis to make the problem a little more clear?
 2 years ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
is it \[\lim_{x \rightarrow 0} \frac{ sinxx }{ x^{3} }\]
 2 years ago

kabulis Group TitleBest ResponseYou've already chosen the best response.0
Yes that is the exact question.. @Eda2012
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Can you use :'hopital's rule?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
L'hopital's rule*
 2 years ago

kabulis Group TitleBest ResponseYou've already chosen the best response.0
Yes, we can use Hopital rule!
 2 years ago

etemplin Group TitleBest ResponseYou've already chosen the best response.0
that makes it so much easier @Dido525 thanks for thinking of that
 2 years ago

Dido525 Group TitleBest 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.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Haha. I always ask ^_^ .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah the limit is 0/0 if you directly sub in.
 2 years ago

kabulis Group TitleBest 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
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1352266897769:dw
 2 years ago

kabulis Group TitleBest ResponseYou've already chosen the best response.0
Thanks @Dido525
 2 years ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
\[1^{st} \] \[\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx1 }{ 3x ^{2} }\] \[2^{nd}\] \[\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ sinx }{ 6x }\] \[3^{rd}\] \[\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ cosx }{ 6 }\] \[=\frac{ 1 }{ 6 }\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah. You often have to use L'hopital's rule multiple times.
 2 years ago

kabulis Group TitleBest 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
 2 years ago

Eda2012 Group TitleBest 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....
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352268277086:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352268373043:dw
 2 years ago

Eda2012 Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012 ...r u using the divergent test...???
 2 years ago
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