Solution:
If f(x) is continuous at $\text{x}=\frac{\pi}{2},$
then$\lim\limits_{\text{x}\rightarrow\frac{\text{x}}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\lim\frac{1-\sin\text{x}}{\text{x}\rightarrow\frac{\pi}{2}(\pi-2\text{x})^2}=\text{f}\Big(\frac{\pi}{2}\Big) \ ...(\text{i})$$$
suppose $\Big(\frac{\pi}{2}-\text{x}\Big)=\text{t},$ then
$\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\sin\Big(\frac{\pi}{2}-\text{t}\Big)}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$ [From eq.(i)]
$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\cos\text{t}}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{4}\lim\limits_{t\rightarrow0}\begin{bmatrix}\frac{2\sin^2\Big(\frac{\text{t}}{2}\Big)}{\text{t}^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow \frac{1}{4}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$
$\Rightarrow \frac{1}{8}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{\sin\Big(\frac{\text{t}}{2}\Big)}{\frac{\text{t}}{2}} \end{bmatrix}^2=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\lambda=\frac{1}{8}$
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