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7.6 Q.II · Q78

Q.lim⁡x→0[3+x3−x]1/x\displaystyle\lim_{x\to 0}\left[\frac{3+x}{3-x}\right]^{1/x}

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✓ Free question

3+x3−x−1=2x3−x\dfrac{3+x}{3-x}-1=\dfrac{2x}{3-x}. So [3+x3−x]1/x={[1+2x3−x](3−x)/(2x)}2/(3−x)\left[\dfrac{3+x}{3-x}\right]^{1/x}=\left\{\left[1+\dfrac{2x}{3-x}\right]^{(3-x)/(2x)}\right\}^{2/(3-x)}. As x→0x\to0, the inner brace →e\to e and the outer exponent →2/3\to2/3. So the limit is e2/3e^{2/3}.

✓Final answer

e2/3e^{2/3}

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