Solve this

Solution:

The function $f(x)=\left\{\begin{array}{cl}\frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ k, & x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$.

$\therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)$

$\Rightarrow k=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\pi-2 x}$

Put $x=\frac{\pi}{2}-h$

When $x \rightarrow \frac{\pi}{2}, h \rightarrow 0$

$\therefore k=\lim _{h \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}$

$\Rightarrow k=\lim _{h \rightarrow 0} \frac{1-\cos h}{2 h}$

$\Rightarrow k=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2}}{2 h}$

$\Rightarrow k=\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{h} \times \lim _{h \rightarrow 0} \sin \frac{h}{2}$

$\Rightarrow k=\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{2 \times \frac{h}{2}} \times \lim _{h \rightarrow 0} \sin \frac{h}{2}$

$\Rightarrow k=\frac{1}{2} \times \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim _{h \rightarrow 0} \sin \frac{h}{2}$

$\Rightarrow k=\frac{1}{2} \times 1 \times 0$

$\Rightarrow k=0$

Thus, the value of k is 0.

If $f(x)=\left\{\begin{array}{cl}\frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ k, & x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then $k=$ ____0____.