Conceptual very.

put a=b,
\lim_{x\rightarrow 0}\frac{a\cos x-a+2cx+3x^2}{4x\log(1+x)+2x^2/(1+x)-6x^2+4x^3}

\lim_{x\rightarrow 0}\frac{-a\sin x+2c+6x}{4\log(1+x)+4x/(1+x)+4x/(1+x) -2x^2/(1+x)^2-12x+12x^2}

\lim_{x\rightarrow 0}\frac{-a\sin x+2c+6x}{4\log(1+x)+8x/(1+x)-2x^2/(1+x)^2-12x+12x^2}

denominator tends to zero so c=0
apply LH rule again...

\lim_{x\rightarrow 0}\frac{-a\cos x+6}{4/(1+x)+8/(1+x)-8x/(1+x)^2-4x/(1+x)^2-4x^2/(1+x)^3-12+24x}

\lim_{x\rightarrow 0}\frac{-a\cos x+6}{12/(1+x)-12x/(1+x)^2-4x^2/(1+x)^3-12+24x}

as x tends to zero , again deno tends to zero .. so a=6

a+b = 12

now you have found the four correct options :)

Then Avni, I think there is something wrong with the question.. let me try to final answer too..

\lim_{x\rightarrow 0}\frac{a\sin x-bx+cx^2+x^3}{2x^2\log(1+x)-2x^3+x^4}

\lim_{x\rightarrow 0}\frac{a\cos x-b+2cx+3x^2}{4x\log(1+x)+2x^2/(1+x)-6x^2+4x^3}

limit x -> zero the denominator tends to zero... numerator tends to a-b

so for limit to exist, a-b has to be zero..

The LH rule is L Hospital rule

it is simply the derivative of numerator to denominator if both of them are zero or infinite..

here see that both the num and deno are zero as x-> 0

so take the derivative of numerator by denominator..

(I assume this is yuor doubt avni?)