anonymous
  • anonymous
determine if existent the limit: lim -->0 sinx-x/x^3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can you put parenthesis to make the problem a little more clear?
anonymous
  • anonymous
is it \[\lim_{x \rightarrow 0} \frac{ sinx-x }{ x^{3} }\]
anonymous
  • anonymous
Yes that is the exact question.. @Eda2012

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Can you use :'hopital's rule?
anonymous
  • anonymous
L'hopital's rule*
anonymous
  • anonymous
Yes, we can use Hopital rule!
anonymous
  • anonymous
that makes it so much easier @Dido525 thanks for thinking of that
anonymous
  • anonymous
Okay then! Take the derivative of the top and bottom terms :) . But before that check if the limit is 0/0 or infinity/ infinity.
anonymous
  • anonymous
Haha. I always ask ^_^ .
anonymous
  • anonymous
Yeah the limit is 0/0 if you directly sub in.
anonymous
  • anonymous
I need to show step by step to the professor :( lol... can anybody do that for me.. sorry guys
anonymous
  • anonymous
|dw:1352266897769:dw|
anonymous
  • anonymous
Thanks @Dido525
anonymous
  • anonymous
Welcome!
anonymous
  • anonymous
\[1^{st} \] \[\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx-1 }{ 3x ^{2} }\] \[2^{nd}\] \[\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ -sinx }{ 6x }\] \[3^{rd}\] \[\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ -cosx }{ 6 }\] \[=-\frac{ 1 }{ 6 }\]
anonymous
  • anonymous
Yeah. You often have to use L'hopital's rule multiple times.
anonymous
  • anonymous
thnx all... what will be the solution if we use the rule L'hopital? since that is the required
anonymous
  • anonymous
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....
anonymous
  • anonymous
|dw:1352268277086:dw|
anonymous
  • anonymous
|dw:1352268373043:dw|
anonymous
  • anonymous
@mahmit2012 ...r u using the divergent test...???

Looking for something else?

Not the answer you are looking for? Search for more explanations.