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eiya
evaluate lim (x-->0) S(x)/(X^3) if S(x)= int (from 0 to x) sin (0.5*pi*t^2)
well, it will be helpful, if u could tell us which grade r u in..?
no...only applying indeterminate forms and L'hospital's Rule chapter
well, may take some time...|dw:1363454097397:dw|
i can see yr hardness to write it....appreciate a lot..!
@Preetha ....posiibility is like she <<can help you.....
Nah. I dont have a clue! Try @klimenkov
how is t related to x ?
actually there "dt" after the close bracket
\[\lim\limits _{x\rightarrow0}\frac{\int\limits _0^x\sin\frac{\pi t^2}2 dt}{x^3}\] 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 \(\frac{\pi}{6}\).
are you applying fundamental theorem?
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 \(\lim\limits_{x\rightarrow0}\cos\frac{\pi x^2}2=1\).
thanks a lot....but im trying to understand the steps..
Do you know the L'hopital's rule?
a bit...just learnt last month
Do you know what is the derivative?
nahh...i got it...understand already....
yes...i know derivative...
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