1708\hspace{-16}\bf{(13)\;}$ Given $\bf{\lim_{x\rightarrow \frac{1}{2}}\frac{\cos^2(\pi x)}{e^{2x}-2e.x}=\frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{2\cos^2 (\pi x)}{e^{2x}-2e.x}}$\\\\\\ $\bf{=\frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{1+\cos (2\pi x)}{e^{2x}-2e.x}\Rightarrow_{Using \; L.Hopital\; Rule} \frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{-2\pi.\sin (2\pi x)}{2e^{2x}-2e}}$\\\\\\ $\bf{=\Rightarrow_{Using \; L.Hopital\; Rule} \frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{-4\pi^2.\cos (2\pi x)}{4e^{2x}}=\frac{1}{2}.\frac{4\pi^2}{4e}=\frac{\pi^2}{2e}}$
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