\lim_{x \to 0} \frac{(e^{3x} - 1) \cdot \ln(x + 1)}{1 - \cos 2x}
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Step 1
Using the double angle formula for cosine, we have: 1 - cos 2x = 1 - (cos^2 x - sin^2 x) = 1 - (1 - 2sin^2 x) = 2sin^2 x Now, let's find the limit as x approaches 0 of 2sin^2 x. As x approaches 0, sin x also approaches 0. Show more…
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