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

Question

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

Accepted Answer

$\begin{gathered} \underset{x	o {1}^{+}}{lim}\frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1{)}^{3}}=-1 \ x-1=t \ \underset{t	o {0}^{+}}{lim}\frac{6t+\lambda t\cos t-\mu \sin t}{{t}^{3}}=-1 \end{gathered}$

$\begin{gathered} =\underset{t	o {0}^{+}}{lim}\frac{6t+\lambda t(1-\frac{{t}^{2}}{2!}+\frac{{t}^{4}}{4!}+\cdots )-\mu (t-\frac{{t}^{3}}{3!}+\cdots )}{{t}^{3}}=-1 \ =\underset{t	o {0}^{+}}{lim}\frac{t(6+\lambda -\mu )+{t}^{3}(-\frac{\lambda }{2}+\frac{\mu }{6})+\cdots }{{t}^{3}}=-1 \end{gathered}$

$\begin{gathered} 	herefore \, \lambda -\mu +6=0\, \ \frac{\mu }{6}-\frac{\lambda }{2}=-1\, \ \ \lambda =6,\mu =12 \ \lambda +\mu =18 \end{gathered}$

Solutions · 1