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Limit Exercise Solved by two methods without using L'Hospital Rule ,


Exercise:

Show that, limx0xarcsinx2xcosxsinx=3

Solution:  Let θ=arcsinx2x2=sinθ

So limx0xarcsinx2xcosxsinx=limx0xθx(cosxsinxx)=limx0θcosxsinxx×sinθsinθ

But limx0θsinθ=limx0arcsinx2x2=limu0arcsinuuH.R=limu011u2=1

So limx0θsinθ(cosxsinxx)sinθ=limx0sinθcosxsinxx=limx0xsinθxcosxsinx=limx0x3xcosxsinx

                                  =limx0x3cosxxtanx=limx0x3secxxtanx=limx0x3xtanx

Let L=limx0xtanxx3

Put x=3u as x0,u0

So L=limu03utan3u27u327L=limu03utan3uu3

Recall that tan3u=tan3u3tanu3tan2u1

Hence 27L=limu03utan3u3tanu3tan2u1u3=limu09utan2u3utan3u+3tanu3tan2u1u3

              =limu09utan2u3utan3u+3tanu3u3tan2uu3

Hence 27L=limu013tan2u1×[limu03(tanuu)u3limu0tan3uu3+limu09utan2uu3]

27L=3limu0tanuuu3+limu0tan3uu39limu0tan2uu2

But limu0tanuuu3=limu0utanuu3=limx0xtanxx3=L

And limu0tanuu=limu0sinucosuu1=limu0sinuucosu=limu0sinuu=1

Thus 27L=3L+1927L3L=824L=8L=13

Hence limx0xtanxx3=1limx0xtanxx3=3

Thus limx0xarcsinx2xcosxsinx=3

2nd method using Series expansion

We know that ddx(arcsinx2)=2x1x4

And th series of expansion of 11x2=(1x2)1/2=i=0(1/2i)(x2)i=1+x22+38x4+1548x6+...,1<x1

2x1x4=2x(1x4)1/2=2x(1+x42+3x88+15x1248+....)    ,  1<x1

Thus arcsinx2=x02x1x4dx=x0(2x+x5+6x98+30x1348)dx=x2+x66+6x1080+30x1448×14+..

Also we have sinx=xx33!+x55!x77!+.....&cosx=1x22!+x44!x66!+...

Hence limx0xarcsinx2xcosxsinx=limx0x(x2+x66+6x1080+...)x(1x22+x424x66!)(xx33!+x55!x77!+...)

                              =limx0x3(1+16x3+680x7+.....)xx32+x524x76!x+x33!x55!+x77!+...

=limx0x3(1+x36+6x780+....)x3(12+13!+x224x25!x46!+x47!)=112+16=16+36=63=3



*_______________________
The idea of First Method Credit to María Jesús Méndez  

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