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7.6 Q.II · Q82

Q.lim⁡x→0[5+7x5−3x]1/3x\displaystyle\lim_{x\to 0}\left[\frac{5+7x}{5-3x}\right]^{1/3x}

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5+7x5−3x−1=10x5−3x\dfrac{5+7x}{5-3x}-1=\dfrac{10x}{5-3x}. The exponent's overall limit is $\lim_{x\to0}\dfrac{1}{3x}\cdot\dfrac{10x}{5-3x}=\dfrac{10}{3\times5}=\ …

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