eiya 2 years ago evaluate lim (x-->0) S(x)/(X^3) if S(x)= int (from 0 to x) sin (0.5*pi*t^2)

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1. Koikkara

well, it will be helpful, if u could tell us which grade r u in..?

2. eiya

univrsity..

3. Koikkara

Fourier series?

4. eiya

no...only applying indeterminate forms and L'hospital's Rule chapter

5. Koikkara

well, may take some time...|dw:1363454097397:dw|

6. eiya

i can see yr hardness to write it....appreciate a lot..!

7. Koikkara

8. Preetha

Nah. I dont have a clue! Try @klimenkov

9. AravindG

how is t related to x ?

10. eiya

actually there "dt" after the close bracket

11. AravindG

thats better :)

12. klimenkov

$\lim\limits _{x\rightarrow0}\frac{\int\limits _0^x\sin\frac{\pi t^2}2 dt}{x^3}$ When $$x\rightarrow0$$, $$\int\limits _0^x\sin\frac{\pi t^2}2 dt\sim\frac{\pi x^3}{6}$$, because $$\sin\frac{\pi t^2}2\sim\frac{\pi t^2}{2}$$, when $$t\rightarrow0$$. So the limit is $$\frac{\pi}{6}$$.

13. eiya

are you applying fundamental theorem?

14. klimenkov

Lol. I have just read what you wrote above about L'hopital's rule. $$\left(\int\limits _0^x\sin\frac{\pi t^2}2 dt\right)'_x=\sin\frac{\pi x^2}2$$ $$\left(\sin\frac{\pi x^2}2\right)'_x=\pi x\cos\frac{\pi x^2}2$$ So we have $$\lim\limits _{x\rightarrow0}\frac{\int\limits _0^x\sin\frac{\pi t^2}2 dt}{x^3}=\lim\limits _{x\rightarrow0}\frac{\sin\frac{\pi x^2}2}{3x^2}=\lim\limits _{x\rightarrow0}\frac{\pi x\cos\frac{\pi x^2}2}{6x}=\lim\limits _{x\rightarrow0}\frac{\pi \cos\frac{\pi x^2}2}{6}=\frac \pi 6$$, because $$\lim\limits_{x\rightarrow0}\cos\frac{\pi x^2}2=1$$.

15. eiya

thanks a lot....but im trying to understand the steps..

16. klimenkov

Do you know the L'hopital's rule?

17. eiya

a bit...just learnt last month

18. klimenkov

Do you know what is the derivative?

19. eiya

20. eiya

yes...i know derivative...

21. eiya

thank you..!

22. klimenkov

You are welcome.

23. mathslover

Welcome to OpenStudy @eiya

24. mathslover

http://prezi.com/fs3hqdpcopic/an-unofficial-guide-to-openstudy/ ^ A guide for you to start your journey well in openstudy @eiya