anonymous
  • anonymous
evaluate lim (x-->0) S(x)/(X^3) if S(x)= int (from 0 to x) sin (0.5*pi*t^2)
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Koikkara
  • Koikkara
well, it will be helpful, if u could tell us which grade r u in..?
anonymous
  • anonymous
univrsity..
Koikkara
  • Koikkara
Fourier series?

Looking for something else?

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

More answers

anonymous
  • anonymous
no...only applying indeterminate forms and L'hospital's Rule chapter
Koikkara
  • Koikkara
well, may take some time...|dw:1363454097397:dw|
anonymous
  • anonymous
i can see yr hardness to write it....appreciate a lot..!
Koikkara
  • Koikkara
@Preetha ....posiibility is like she <
Preetha
  • Preetha
Nah. I dont have a clue! Try @klimenkov
AravindG
  • AravindG
how is t related to x ?
anonymous
  • anonymous
actually there "dt" after the close bracket
AravindG
  • AravindG
thats better :)
klimenkov
  • 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 \(\frac{\pi}{6}\).
anonymous
  • anonymous
are you applying fundamental theorem?
klimenkov
  • 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 \(\lim\limits_{x\rightarrow0}\cos\frac{\pi x^2}2=1\).
anonymous
  • anonymous
thanks a lot....but im trying to understand the steps..
klimenkov
  • klimenkov
Do you know the L'hopital's rule?
anonymous
  • anonymous
a bit...just learnt last month
klimenkov
  • klimenkov
Do you know what is the derivative?
anonymous
  • anonymous
nahh...i got it...understand already....
anonymous
  • anonymous
yes...i know derivative...
anonymous
  • anonymous
thank you..!
klimenkov
  • klimenkov
You are welcome.
mathslover
  • mathslover
Welcome to OpenStudy @eiya
mathslover
  • mathslover
http://prezi.com/fs3hqdpcopic/an-unofficial-guide-to-openstudy/ ^ A guide for you to start your journey well in openstudy @eiya

Looking for something else?

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