# How do you find the limit of #[(1/ln(x+1)) - (1/x)]# as x approaches 0?

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To find the limit of [(1/ln(x+1)) - (1/x)] as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get [(1/(x+1)) - (1/x^2)] / (1/(x+1)). Simplifying this expression, we have [(x^2 - (x+1)) / (x(x+1))] / (1/(x+1)). Further simplifying, we get [(x^2 - x - 1) / x] / (1/(x+1)). Now, substituting x = 0 into this expression, we have [(0^2 - 0 - 1) / 0] / (1/(0+1)). Simplifying further, we get (-1/0) / (1/1), which is undefined. Therefore, the limit does not exist.

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To find the limit of ((\frac{1}{\ln(x+1)}) - \frac{1}{x}) as (x) approaches 0, first, notice that as (x) approaches 0, (\ln(x+1)) approaches 0, and (\frac{1}{x}) approaches positive or negative infinity depending on whether (x) approaches 0 from the right or left. Therefore, the limit is undefined.

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