Here's the question you clicked on:
kabulis
determine if existent the limit: lim -->0 sinx-x/x^3
can you put parenthesis to make the problem a little more clear?
is it \[\lim_{x \rightarrow 0} \frac{ sinx-x }{ x^{3} }\]
Yes that is the exact question.. @Eda2012
Can you use :'hopital's rule?
Yes, we can use Hopital rule!
that makes it so much easier @Dido525 thanks for thinking of that
Okay then! Take the derivative of the top and bottom terms :) . But before that check if the limit is 0/0 or infinity/ infinity.
Haha. I always ask ^_^ .
Yeah the limit is 0/0 if you directly sub in.
I need to show step by step to the professor :( lol... can anybody do that for me.. sorry guys
\[1^{st} \] \[\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx-1 }{ 3x ^{2} }\] \[2^{nd}\] \[\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ -sinx }{ 6x }\] \[3^{rd}\] \[\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ -cosx }{ 6 }\] \[=-\frac{ 1 }{ 6 }\]
Yeah. You often have to use L'hopital's rule multiple times.
thnx all... what will be the solution if we use the rule L'hopital? since that is the required
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....
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@mahmit2012 ...r u using the divergent test...???