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Miscellaneous 7 II · Q121

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

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For small xx, ex≈1+x+x2/2e^x\approx1+x+x^2/2 and e−x≈1−x+x2/2e^{-x}\approx1-x+x^2/2; adding, ex+e−x−2≈x2e^x+e^{-x}-2\approx x^2. The denominator $x\tan x\sim x\cdot x= …

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