LIMITS again

Limit = \lim_{x\rightarrow 0}\frac{a(e^{x}+xe^{x})-b.\frac{1}{1+x}+c(e^{-x}-xe^{-x})}{2xsinx+x^{2}cosx} [ using L H rule ]
= a - b + c0 = \infty unless a - b + c =0
but given limit is 2 so inderterminate form 0/0 shuld continue
so a - b + c = 0 .............................(1)
then limit , [ using LH rule ]

= \lim_{x\rightarrow 0} \frac{\left\{a(1+x)e^{x}+ae^{x}+\frac{b}{(1+x)^{2}}+c(-1)e^{-x} +c(1-x)e^{-x}.(-1)\right\}}{2sinx+2xcoxsx+2xcosx-x^{2}sinx}

= \lim_{x\rightarrow 0} \frac{\left\{a(2+x)e^{x}+\frac{b}{(1+x)^{2}} -c(2-x)e^{-x}\right\}}{2sinx+4xcoxsx-x^{2}sinx}

= \frac{2a+b-2c}0{} = \infty unless 2a + b - 2c = 0
but limit = 2
so, 2a + b - 2c = 0 .............................(2)

then limit = \lim_{x\rightarrow 0} \frac{\left\{a(2+x)e^{x}+ae^{x}-\frac{2b}{(1+x)^{3}}+c(2-x)e^{-x}+ce^{-x} \right\}}{6cosx-4xsinx-2xsinx-x^{2}cosx}

= 2a + a - 2b + 2c + c6 = 2 (given )
hence 3a - 2b + 3c = 12 .............................(3)

solving equation (1) , (2) and (3) we get
a = 3 , b = 12 , c = 9 .