If $\lim _\limits{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda+\mu$ is equal to

<p>$$\begin{aligned} &\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\\ &\text { Let } x-1=t\\ &\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t \cos t-\mu \sin t}{t^3}=-1 \end{aligned}$$</p> <p>$$\begin{aligned} & =\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t\left(1-\frac{t^2}{2!}+\frac{t^4}{4!}+\cdots\right)-\mu\left(t-\frac{t^3}{3!}+\cdots\right)}{t^3}=-1 \\ & =\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda-\mu)+t^3\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)+\cdots}{t^3}=-1 \end{aligned}$$</p> <p>$$\begin{aligned} &\begin{aligned} \therefore\quad & \lambda-\mu+6=0 \quad\text{..... (i)}\\ & \frac{\mu}{6}-\frac{\lambda}{2}=-1 \quad\text{..... (ii)} \end{aligned}\\ &\text { Solving (i) and (ii) }\\ &\begin{aligned} & \lambda=6, \mu=12 \\ & \lambda+\mu=18 \end{aligned} \end{aligned}$$</p>