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Miscellaneous 7 II · Q125

Q.lim⁡x→2[log⁡x−log⁡2x−2]\displaystyle\lim_{x\to 2}\left[\frac{\log x-\log 2}{x-2}\right]

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Let t=x−2→0t=x-2\to0, so x=2+tx=2+t. log⁡x−log⁡2=log⁡(2+t)−log⁡2=log⁡ ⁣(1+t2)\log x-\log2=\log(2+t)-\log2=\log\!\left(1+\dfrac t2\right). Dividing by tt: $\dfrac{\log(1+t/2)}{t}=\dfrac12\cdot\dfrac{\log( …

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