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If $\lim_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$ exists and is equal to $b$, then the value of $a - 2b$ is
Applying L’Hospital Rule twice: $a = 4$. Then $b = \lim_{x\to 0} \frac{-16}{6a + 6a} = \frac{-16}{48} = -\frac{1}{3}$. So $a - 2b = 4 - 2(-\frac{1}{3}) = 4 + 1 = 5$.
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