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7.6 Q.I · Q72

Q.lim⁡x→0[9x−5x4x−1]\displaystyle\lim_{x\to 0}\left[\frac{9^x-5^x}{4^x-1}\right]

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✓ Free question

Divide numerator and denominator by xx: [(9x−1)−(5x−1)]/x(4x−1)/x→log⁡9−log⁡5log⁡4=log⁡(9/5)log⁡4\dfrac{[(9^x-1)-(5^x-1)]/x}{(4^x-1)/x}\to\dfrac{\log9-\log5}{\log4}=\dfrac{\log(9/5)}{\log4}.

✓Final answer

log⁡(9/5)log⁡4\dfrac{\log(9/5)}{\log4}

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