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Miscellaneous 7 I · Q108

Q.lim⁡x→0[log⁡(5+x)−log⁡(5−x)sin⁡x]=\displaystyle\lim_{x\to 0}\left[\frac{\log(5+x)-\log(5-x)}{\sin x}\right]= (A) 32\dfrac{3}{2} (B) −52-\dfrac{5}{2} (C) −12-\dfrac{1}{2} (D) 25\dfrac{2}{5}

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log⁡(5+x)−log⁡(5−x)=log⁡(1+x/5)−log⁡(1−x/5)\log(5+x)-\log(5-x)=\log(1+x/5)-\log(1-x/5) (the log⁡5\log5 terms cancel). Dividing by xx: 15−(−15)=25\dfrac15-\left(-\dfrac15\right)=\dfrac25. Multiplying by $x/\sin x\t …

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