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7.6 Q.II · Q77

Q.lim⁡x→0[3x+3−x−2x⋅tan⁡x]\displaystyle\lim_{x\to 0}\left[\frac{3^x+3^{-x}-2}{x\cdot\tan x}\right]

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Using the series-style expansion (or the identity (3x/2−3−x/2)2=3x−2+3−x(3^{x/2}-3^{-x/2})^2=3^x-2+3^{-x}), for small xx, 3x+3−x−2∼x2(log⁡3)23^x+3^{-x}-2\sim x^2(\log3)^2. The denominator xtan⁡x∼x2x\tan x\sim x^2. So the limit is x2(log⁡3)2x2=(log⁡3)2\dfrac{x^2(\log3)^2}{x^2}=(\log3)^2.

✓Final answer

(log⁡3)2(\log3)^2

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