Question

Solve : ${lim}_{x\to 0}\frac{\sin 2x+a\sin x}{x^3}=p\left(\text{finite}\right)$

Answer 1

$\begin{array}{c}Let\\ L={lim}_{x\to 0}\frac{\sin 2x+a\sin x}{x^3}\\ ={lim}_{x\to 0}\frac{\sin x\left(2\cos x+a\right)}{x\times x^2}\\ ={lim}_{x\to 0}\frac{2\cos x+a}{x^2}\\ LetLiffinite\\ {lim}_{x\to 0}{D}^{r}iszero\\ \therefore Lisfinite\\ here\\ {lim}_{x\to 0}{N}^r=0\\ \Rightarrow {lim}_{x\to 0}2\cos x+a=0\\ \Rightarrow 2+a=0\\ \therefore a=-2\\ Now\\ L={lim}_{x\to 0}\frac{2\cos x-2}{x^2}\\ =-2{lim}_{x\to 0}\frac{1-\cos x}{x^2}\\ =-2{lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}\\ =\frac{-4}{4}{lim}_{x\to 0}\frac{{\sin }^2\frac{x}{2}}{{\left(\frac{x}{2}\right)}^2}\\ =-1{lim}_{x\to 0}{\left(\frac{\sin \frac{x}{2}}{\left(\frac{x}{2}\right)}\right)}^2\\ =-1\times 1^2\left[{lim}_{x\to 0}\frac{\sin x}{x}=1\right]\\ =-1\end{array}$