Let f(x) = \ cases ( 1 - x (2π - x) ^ 2 )( ^ 2 x (1 + 4 π ^ 2 - 4 +...

For the function $f$\left(x\right)$to be continuous at$x = 2\pi$<br /><br />$\lim_{x \to 2\pi}f\left(x\right) = f\left(2\pi\right)$<br /><br />Now,$\lim_{x \to 2\pi}\dfrac{1 - cos x}{{\left(2\pi - x\right)}^{2}}\dfrac{si{n}^{2}x}{\log \left\{1 + {\left(2\pi - x\right)}^{2}\right\}} = \lambda$<br /><br />Putting$x = 2\pi + t$, we get, <br /><br />$\lim_{t \to 0}\dfrac{1 - \cos t}{{t}^{2}} ⋅ \dfrac{si{n}^{2}t}{\log \left(1 + {t}^{2}\right)} = \lambda$<br /><br />$\lim_{t \to 0}\dfrac{1}{2} ⋅ \dfrac{si{n}^{2}t}{{t}^{2}} ⋅ \dfrac{{t}^{2}}{\log \left(1 + {t}^{2}\right)} = \lambda$<br /><br />$\dfrac{1}{2} \times 1 \times 1 = \lambda ⇒ \lambda = \dfrac{1}{2}