Mathematics - Limits Question with Solution | TestHub

: ${\mathrm{Lim}}_{\mathrm{x}\to \infty }\frac{{\tan }^{-1}\left(\frac{\mathrm{x}+\lambda }{\mathrm{x}+\mu }\right)-\frac{\pi }{4}}{\frac{1}{\mathrm{x}}}=1$ Apply L'Hopital's rule and simplifying, we get

$$\begin{array}{rl}& {\mathrm{Lim}}_{\mathrm{x}\to \infty }\frac{(\lambda -\mu ){\mathrm{x}}^2}{2{\mathrm{x}}^2+2\mathrm{x}(\lambda +\mu )+\left({\mu }^2+{\lambda }^2\right)}=1\\ & \Rightarrow \frac{\lambda -\mu }{2}=1\\ & \Rightarrow \lambda -\mu =2\\ & \therefore (\lambda ,\mu )\text{ can be }(5,3)\end{array}$$