If the function [f(x)=cases(1+| x|)^/| x|, & 0

Question

If the function
$$f(x)=\begin{cases}(1+|\cos x|)^{\lambda/|\cos x|}, & 0<x<\dfrac\pi2\ \mu, & x=\dfrac\pi2\ e^{\cot6x/\cot4x}, & \dfrac\pi2<x<\pi\end{cases}$$
is continuous at $x=\dfrac\pi2$, then $9\lambda+6\log_e\mu+\mu^6-e^{6\lambda}$ is equal to

Accepted Answer

Right-hand limit: $\displaystyle\lim_{x\to\pi/2^+}e^{\cot6x/\cot4x}=\lim_{x\to\pi/2^+}e^{\frac{\sin4x\cos6x}{\sin6x\cos4x}}=e^{2/3}$ (evaluating the ratio of derivatives/limits at $x=\pi/2$).

Left-hand limit: $\displaystyle\lim_{x\to\pi/2^-}(1+|\cos x|)^{\lambda/|\cos x|}=e^{\lambda}$ (since $(1+u)^{1/u}\to e$ as $u\to0$).

For continuity, $e^{2/3}=e^\lambda=\mu$, so $\lambda=\dfrac23,\ \mu=e^{2/3}$.

$$9\lambda+6\log_e\mu+\mu^6-e^{6\lambda}=9\cdot\frac23+6\cdot\frac23+\left(e^{2/3}\right)^6-e^{6\cdot2/3}=6+4+e^4-e^4=10$$

Solutions · 1