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

1. etemplin Group Title

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

2. Eda2012 Group Title

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

3. kabulis Group Title

Yes that is the exact question.. @Eda2012

4. Dido525 Group Title

Can you use :'hopital's rule?

5. Dido525 Group Title

L'hopital's rule*

6. kabulis Group Title

Yes, we can use Hopital rule!

7. etemplin Group Title

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

8. Dido525 Group Title

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 Group Title

Haha. I always ask ^_^ .

10. Dido525 Group Title

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

11. kabulis Group Title

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

12. Dido525 Group Title

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13. kabulis Group Title

Thanks @Dido525

14. Dido525 Group Title

Welcome!

15. Eda2012 Group Title

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

16. Dido525 Group Title

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

17. kabulis Group Title

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

18. Eda2012 Group Title

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 Group Title

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20. mahmit2012 Group Title

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21. Eda2012 Group Title

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