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7.6 Q.II · Q81

Q.lim⁡x→0[4x+11−4x]1/x\displaystyle\lim_{x\to 0}\left[\frac{4x+1}{1-4x}\right]^{1/x}

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4x+11−4x−1=8x1−4x\dfrac{4x+1}{1-4x}-1=\dfrac{8x}{1-4x}. The exponent's overall limit is $\lim_{x\to0}\dfrac1x\cdot\dfrac{8x}{1-4x}=\dfra …

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