Exercise:
Show that, limx→0xarcsinx2xcosx−sinx=−3
Solution: Let θ=arcsinx2⇔x2=sinθ
So limx→0xarcsinx2xcosx−sinx=limx→0xθx(cosx−sinxx)=limx→0θcosx−sinxx×sinθsinθ
But limx→0θsinθ=limx→0arcsinx2x2=limu→0arcsinuuH.R=limu→01√1−u2=1
So limx→0θsinθ(cosx−sinxx)sinθ=limx→0sinθcosx−sinxx=limx→0xsinθxcosx−sinx=limx→0x3xcosx−sinx
=limx→0x3cosxx−tanx=limx→0x3secxx−tanx=limx→0x3x−tanx
Let L=limx→0x−tanxx3
Put x=3u as x→0,u→0
So L=limu→03u−tan3u27u3⇔27L=limu→03u−tan3uu3
Recall that tan3u=tan3u−3tanu3tan2u−1
Hence 27L=limu→03u−tan3u−3tanu3tan2u−1u3=limu→09utan2u−3u−tan3u+3tanu3tan2u−1u3
=limu→09utan2u−3u−tan3u+3tanu3u3tan2u−u3
Hence 27L=limu→013tan2u−1×[limu→03(tanu−u)u3−limu→0tan3uu3+limu→09utan2uu3]
⇒27L=−3limu→0tanu−uu3+limu→0tan3uu3−9limu→0tan2uu2
But limu→0tanu−uu3=−limu→0u−tanuu3=−limx→0x−tanxx3=−L
And limu→0tanuu=limu→0sinucosuu1=limu→0sinuucosu=limu→0sinuu=1
Thus 27L=3L+1−9⇔27L−3L=−8⇔24L=−8⇔L=−13
Hence limx→0x−tanxx3=1limx→0x−tanxx3=−3
Thus limx→0xarcsinx2xcosx−sinx=−3
2nd method using Series expansion
We know that ddx(arcsinx2)=2x√1−x4
And th series of expansion of 1√1−x2=(1−x2)−1/2=∞∑i=0(−1/2i)(−x2)i=1+x22+38x4+1548x6+...,−1<x≤1
⇒2x√1−x4=2x(1−x4)−1/2=2x(1+x42+3x88+15x1248+....) , −1<x≤1
Thus arcsinx2=∫x02x√1−x4dx=∫x0(2x+x5+6x98+30x1348)dx=x2+x66+6x1080+30x1448×14+..
Also we have sinx=x−x33!+x55!−x77!+.....&cosx=1−x22!+x44!−x66!+...
Hence limx→0xarcsinx2xcosx−sinx=limx→0x(x2+x66+6x1080+...)x(1−x22+x424−x66!)−(x−x33!+x55!−x77!+...)
=limx→0x3(1+16x3+680x7+.....)x−x32+x524−x76!−x+x33!−x55!+x77!+...
=limx→0x3(1+x36+6x780+....)x3(−12+13!+x224−x25!−x46!+x47!)=1−12+16=1−6+36=6−3=−3
*_______________________
The idea of First Method Credit to María Jesús Méndez
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