## kabulis determine if existent the limit: lim -->0 sinx-x/x^3 one year ago one year ago

1. etemplin

can you put parenthesis to make the problem a little more clear?

2. Eda2012

is it $\lim_{x \rightarrow 0} \frac{ sinx-x }{ x^{3} }$

3. kabulis

Yes that is the exact question.. @Eda2012

4. Dido525

Can you use :'hopital's rule?

5. Dido525

L'hopital's rule*

6. kabulis

Yes, we can use Hopital rule!

7. etemplin

that makes it so much easier @Dido525 thanks for thinking of that

8. Dido525

Okay then! Take the derivative of the top and bottom terms :) . But before that check if the limit is 0/0 or infinity/ infinity.

9. Dido525

Haha. I always ask ^_^ .

10. Dido525

Yeah the limit is 0/0 if you directly sub in.

11. kabulis

I need to show step by step to the professor :( lol... can anybody do that for me.. sorry guys

12. Dido525

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13. kabulis

Thanks @Dido525

14. Dido525

Welcome!

15. Eda2012

$1^{st}$ $\frac{ dy }{ dx }= \lim_{x \rightarrow 0}\frac{ cosx-1 }{ 3x ^{2} }$ $2^{nd}$ $\frac{ dy }{ dx } = \lim_{x \rightarrow 0}\frac{ -sinx }{ 6x }$ $3^{rd}$ $\frac{ dy }{ dx }=\lim_{x \rightarrow 0}\frac{ -cosx }{ 6 }$ $=-\frac{ 1 }{ 6 }$

16. Dido525

Yeah. You often have to use L'hopital's rule multiple times.

17. kabulis

thnx all... what will be the solution if we use the rule L'hopital? since that is the required

18. Eda2012

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....

19. mahmit2012

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20. mahmit2012

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21. Eda2012

@mahmit2012 ...r u using the divergent test...???