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Miscellaneous 7 II · Q133

Q.lim⁡x→1[x−1log⁡x]\displaystyle\lim_{x\to 1}\left[\frac{\sqrt{x}-1}{\log x}\right]

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Let u=x−1→0u=x-1\to0, so x=1+ux=1+u. x−1=1+u−1∼u2\sqrt x-1=\sqrt{1+u}-1\sim\dfrac{u}{2} (standard fractional-power approximation, p=1/2p=1/2). log⁡x=log⁡(1+u)∼u\log x=\log(1+u)\sim u. Th …

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