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Question Number 183927 by greougoury555 last updated on 31/Dec/22

$$\underset{x\to 2}{\lim }\frac{\tan \left(2\pi x\right)+\cos \left(\frac{\pi }{2}x\right)+\tan \left(\frac{\pi }{8}x\right)}{x^2+4x-12}=?$$

Answered by cortano1 last updated on 31/Dec/22

$$\underset{x\to 2}{\lim }\frac{\tan \left(2\pi x\right)+\cos \left(\frac{\pi }{2}x\right)+\tan \left(\frac{\pi }{8}x\right)}{(x-2)(x+6)}$$$$=\underset{x\to 2}{\lim }\frac{\tan \left(2\pi x\right)+\cos \left(\frac{\pi }{2}x\right)+\tan \left(\frac{\pi }{8}x\right)}{8(x-2)}$$$$[x-2=m]$$$$=\underset{m\to 0}{\lim }\frac{\tan \left(2\pi (m+2)\right)+\cos \left(\frac{\pi }{2}(m+2)\right)+\tan \left(\frac{\pi }{8}(m+2)\right)}{8m}$$$$=\underset{m\to 0}{\lim }\frac{\tan \left(2\pi m\right)-\cos \left(\frac{\pi }{2}m\right)+\tan (\frac{\pi }{8}m+\frac{\pi }{4})}{8m}$$$$=\frac{\pi }{4}+\underset{m\to 0}{\lim }\frac{1-\cos \left(\frac{\pi }{2}m\right)}{8m}+\underset{m\to 0}{\lim }\frac{\tan (\frac{\pi }{8}m+\frac{\pi }{4})-\tan \frac{\pi }{4}}{8m}$$$$=\frac{\pi }{4}+\underset{m\to 0}{\lim }\frac{\sin {}^2\left(\frac{\pi }{2}m\right)}{16m}+\underset{m\to 0}{\lim }\frac{\tan \left(\frac{\pi }{8}m\right)[1+\tan (\frac{\pi }{8}m+\frac{\pi }{4})]}{8m}$$$$=\frac{\pi }{4}+0+\frac{2.\pi }{64}=\frac{9\pi }{32}$$

Answered by qaz last updated on 31/Dec/22

$$\underset{x\to 2}{lim}\frac{\tan \left(2\pi x\right)+\cos \left(\frac{\pi }{2}x\right)+\tan \left(\frac{\pi }{8}x\right)}{x^2+4x-12}$$$$=\underset{x\to 2}{lim}\frac{2\pi \sec {}^2\left(2\pi x\right)-\frac{\pi }{2}\sin \left(\frac{\pi }{2}x\right)+\frac{\pi }{8}\sec {}^2\left(\frac{\pi }{8}x\right)}{2x+4}$$$$=\frac{2\pi -0+\frac{\pi }{8}\cdot 2}{8}=\frac{9}{32}\pi$$

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