If $\lim _{x \rightarrow 0} \frac{a \sin x-b x+c x^2+x^3}{2 x^2 \ln (1+x)-2 x^3+x^4}$ exists and is

Question

If $\lim _{x \rightarrow 0} \frac{a \sin x-b x+c x^2+x^3}{2 x^2 \ln (1+x)-2 x^3+x^4}$ exists and is equal to $l$ then $a+b+c+l=$

Professor PC Thomas Classes · Answer

Given limit $=\lim _{x ightarrow 0} \frac{a\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots \ldotsight)-b x+c x^2+x^3}{2 x^2\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots . .ight)-2 x^3+x^4}$ $$ =\lim _{x ightarrow 0} \frac{(a-b) x+c x^2+\left(1-\frac{a}{6}ight) x^3+\frac{a x^5}{120} \ldots}{2 \frac{x^5}{3}-\frac{x^6}{2}+\ldots} $$ For this limit to exist, we must have $a=b, c=0, a=6$ and given limit $=\frac{a}{120} 	imes \frac{3}{2}=\frac{6 	imes 3}{120 	imes 2}=\frac{3}{40}$.