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Question Number 193532 by horsebrand11 last updated on 16/Jun/23

$$\underset{x\to 2}{\lim }\frac{1-\cos \pi \mathrm{x}}{{(2-\mathrm{x})}^2}=?$$

Answered by aba last updated on 16/Jun/23

$$\mathrm{let}\mathrm{t}=2-\mathrm{x}$$$$\underset{x\to 2}{\lim }\frac{1-\cos \pi \mathrm{x}}{{(2-\mathrm{x})}^2}=\underset{\mathrm{t}\to 0}{\lim }\frac{1-\cos \pi (2+\mathrm{t})}{{\mathrm{t}}^2}=\underset{\mathrm{t}\to 0}{\lim }\frac{1-\cos \pi \mathrm{t}}{{\left(\pi \mathrm{t}\right)}^2}\times {\pi }^2=\frac{{\pi }^2}{2}$$

Answered by cortano12 last updated on 16/Jun/23

$$\mathrm{by}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\mathrm{L}=\underset{x\to 2}{\lim }\frac{\pi \sin \pi \mathrm{x}}{-2(2-\mathrm{x})}=\underset{x\to 2}{\lim }\frac{{\pi }^2\cos \pi \mathrm{x}}{2}$$$$=\frac{{\pi }^2}{2}$$

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