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7.6 Q.II · Q79

Q.lim⁡x→0[5x+33−2x]2/x\displaystyle\lim_{x\to 0}\left[\frac{5x+3}{3-2x}\right]^{2/x}

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5x+33−2x−1=7x3−2x\dfrac{5x+3}{3-2x}-1=\dfrac{7x}{3-2x}. So [5x+33−2x]2/x={[1+7x3−2x](3−2x)/(7x)}14/[x⋅(3−2x)/x]\left[\dfrac{5x+3}{3-2x}\right]^{2/x}=\left\{\left[1+\dfrac{7x}{3-2x}\right]^{(3-2x)/(7x)}\right\}^{14/[x\cdot(3-2x)/x]}; more directly, the exponent's limit is $\lim_{x\to0}\dfr …

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