lim-x-2-tan-2pix-cos-pi-2-x-tan-pi-8-x-x-2-4x-12-

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

\lim_{x \to 2} \frac{\tan (2 π x) + \cos (\frac{π}{2} x) + \tan (\frac{π}{8} x)}{x^{2} + 4 x - 12} = ?

Answered by cortano1 last updated on 31/Dec/22

\lim_{x \to 2} \frac{\tan (2 π x) + \cos (\frac{π}{2} x) + \tan (\frac{π}{8} x)}{(x - 2) (x + 6)}

= \lim_{x \to 2} \frac{\tan (2 π x) + \cos (\frac{π}{2} x) + \tan (\frac{π}{8} x)}{8 (x - 2)}

[x - 2 = m]

= \lim_{m \to 0} \frac{\tan (2 π (m + 2)) + \cos (\frac{π}{2} (m + 2)) + \tan (\frac{π}{8} (m + 2))}{8 m}

= \lim_{m \to 0} \frac{\tan (2 π m) - \cos (\frac{π}{2} m) + \tan (\frac{π}{8} m + \frac{π}{4})}{8 m}

= \frac{π}{4} + \lim_{m \to 0} \frac{1 - \cos (\frac{π}{2} m)}{8 m} + \lim_{m \to 0} \frac{\tan (\frac{π}{8} m + \frac{π}{4}) - \tan \frac{π}{4}}{8 m}

= \frac{π}{4} + \lim_{m \to 0} \frac{\sin^{2} (\frac{π}{2} m)}{16 m} + \lim_{m \to 0} \frac{\tan (\frac{π}{8} m) [1 + \tan (\frac{π}{8} m + \frac{π}{4})]}{8 m}

= \frac{π}{4} + 0 + \frac{2 . π}{64} = \frac{9 π}{32}

Answered by qaz last updated on 31/Dec/22

\underset{x \to 2}{l i m} \frac{\tan (2 π x) + \cos (\frac{π}{2} x) + \tan (\frac{π}{8} x)}{x^{2} + 4 x - 12}

= \underset{x \to 2}{l i m} \frac{2 π \sec^{2} (2 π x) - \frac{π}{2} \sin (\frac{π}{2} x) + \frac{π}{8} \sec^{2} (\frac{π}{8} x)}{2 x + 4}

= \frac{2 π - 0 + \frac{π}{8} \cdot 2}{8} = \frac{9}{32} π

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