If the function f(x) = \ cases 2 + cos x - 1 (π - x) ^ 2 , & x ≠ π...

Given f\left(x\right) = \left\{\begin{cases}\dfrac{\sqrt{2 + cos x} - 1}{{\left(\pi - x\right)}^{2}}, & x \ne \pi \\ k, & x = \pi\end{cases}\right) is continuous atx = \pi$</p><p>Hence,$f\left(\pi\right) = \lim_{x \to \pi}f\left(x\right)$</p><p>$⇒ k = \lim_{x \to \pi}\dfrac{\sqrt{2 + \cos x} - 1}{{\left(\pi - x\right)}^{2}}$</p><p>Consider$L . H . L . = \lim_{x \to {\pi}^{-}}\dfrac{\sqrt{2 + cos x} - 1}{{\left(\pi - x\right)}^{2}}$</p><p>$\lim_{h \to 0}\dfrac{\sqrt{2 + \cos ( \pi - h )} - 1}{{\left(\pi - \left(\pi - h\right)\right)}^{2}}$<br /><br />Also,$\cos \left(\pi - h\right) = - cosh$<br /><br />$⇒ k = \lim_{h \to 0}\dfrac{\sqrt{2 - cosh} - 1}{{h}^{2}}$</p><p>$⇒ k = \lim_{h \to 0}\left(\dfrac{\sqrt{2 - cosh} - 1}{{h}^{2}}\right) \times \left(\dfrac{\sqrt{2 - cosh} + 1}{\sqrt{2 - cosh} + 1}\right)$</p><p>$⇒ k=\lim_{h \to 0}\dfrac{\left(2 - cosh\right) - 1}{{h}^{2}\left\{\sqrt{2 - cosh} + 1\right\}}$</p><p>$⇒ k=\lim_{h \to 0}\left(\dfrac{1 - cosh}{{h}^{2}}\right)\lim_{h \to 0}\dfrac{1}{\left\{\sqrt{2 - cosh} + 1\right\}}$</p><p>Now, using$\cos 2x = 1 - 2\sin^{2}x, ⇒ 1 - \cos 2x = 2\sin^{2}x,$</p><p>$⇒ k = \lim_{h \to 0}\dfrac{2\sin^{2}\left(\dfrac{h}{2}\right)}{4{\left(\dfrac{h}{2}\right)}^{2}} \times \dfrac{1}{\left(\sqrt{2 - 1} + 1\right)}$</p><p>$⇒ k = \dfrac{1}{2} \times \dfrac{1}{2}$</p><p>$⇒ k = \dfrac{1}{4} .$