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Exercise

lim⁡x→0(ln⁡(1+x1−x−2x)x−sin⁡(x))\lim_{x\to0}\left(\frac{\ln\left(\frac{1+x}{1-x}-2x\right)}{x-\sin\left(x\right)}\right)

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limx→0(ln(1+x1−x −2x)x−sin(x) )
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