If _x 1^+ ((x-1)(6+ (x-1))+ (1-x))/((x-1)^3)=-1, where , R, then + is equal to

Question

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

Accepted Answer

$$\begin{aligned}
&\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\
&\text { Let } x-1=t\
&\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t \cos t-\mu \sin t}{t^3}=-1
\end{aligned}$$

$$\begin{aligned}
& =\lim _{t \rightarrow 0^{+}} \frac{6 t+\lambda t\left(1-\frac{t^2}{2!}+\frac{t^4}{4!}+\cdots\right)-\mu\left(t-\frac{t^3}{3!}+\cdots\right)}{t^3}=-1 \
& =\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda-\mu)+t^3\left(-\frac{\lambda}{2}+\frac{\mu}{6}\right)+\cdots}{t^3}=-1
\end{aligned}$$

$$\begin{aligned}
&\begin{aligned}
\therefore\quad & \lambda-\mu+6=0 \quad\text{..... (i)}\
& \frac{\mu}{6}-\frac{\lambda}{2}=-1 \quad\text{..... (ii)}
\end{aligned}\
\&\text { Solving (i) and (ii) }\
\&\begin{aligned}
& \lambda=6, \mu=12 \
& \lambda+\mu=18
\end{aligned}
\end{aligned}$$

Solutions · 1