Question

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

Answer 1

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

$$\begin{gathered} \therefore f\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right) \\ \Rightarrow k=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{1-\sin x}{\mathrm{\pi }-2x} \end{gathered}$$

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

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

$\therefore k=\underset{h\to 0}{\lim }\frac{1-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{1-\cos h}{2h}$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{2h}$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{h}\times \underset{h\to 0}{\lim }\sin \frac{h}{2}$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{2\times {\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\sin \frac{h}{2}$

$\Rightarrow k=\frac{1}{2}\times \underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\sin \frac{h}{2}$
$\Rightarrow k=\frac{1}{2}\times 1\times 0$

$\Rightarrow k=0$

Thus, the value of k is 0.



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