If $L=\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x+a e^{x}+b e^{-x}+c \ln (1+x)}{x^{3}}=\infty$

Equation $a x^{2}+b x+c=0$ has

Given :- $L=\lim _{x \rightarrow 0} \frac{\operatorname{Sin}+a e^{x}+b e^{-x}+c \ln (1+x)}{x^{3}}=\infty$

To find: - Roots of equation $a x^{2}+b x+c=0$

$\lim _{x \rightarrow 0} \frac{\sin x+a e^{x}+b e^{-x}+c \ln (1+x)}{x^{3}}=\infty$

Applying La Hospital Rule

$\begin{array}{l} \lim _{x \rightarrow 0}\left(x-\frac{x^{3}}{3!}\right)+a\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}\right)+b\left(1-\frac{x}{1!}+\frac{x^{2}}{2!}\right. \\ \left.-\frac{x^{3}}{3!}\right)+c\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}\right) \\ \times \frac{1}{x^{3}} \\ =\lim _{x \rightarrow 0}(a+b)+(1+a-b+c) x+\left(\frac{a}{2}+\frac{b}{2}-\frac{c}{2}\right) x^{2}+\left(\frac{-1}{3!}+\frac{a}{3!}\right. \\ \left.\frac{-b}{3!}+\frac{c}{3}\right) \times \frac{1}{x^{3}} \\ a+b=0 \quad-\text { (1) } \quad 1+a-b+c=0-(2) \quad \frac{a}{2}+\frac{b}{2}-\frac{c}{2}=0 \\ \end{array}$

Eq (3) can be written as

$a+b=c$

Comparing (1) and (4)

$C=0$

Put ( $x$ ) in

$\begin{array}{l} 1+a-b+0=0 \\ \Rightarrow a-b=-1 \end{array}$

Subteacting (1) and (5)

$\begin{array}{l} \text { Put }(* *) \text { in (1) } \\ a+b=0 \\ -\frac{1}{2}+b=0 \Rightarrow b=\frac{1}{2}-(* *) \\ a x^{2}+b x+c=0 \\ \Rightarrow \frac{-1}{2} x^{2}+\frac{1}{2} x+c=0 \\ \Rightarrow x^{2}-x=0 \\ D=b^{2}-4 a c=1-4 \times 1 \times 0=1 \\ 0>0 \text { real roots } \\ x(x-1)=0 \\ x=0,1 \\ \end{array}$