# How do you find the limit of #x^(2x)# as x approaches 0?

Applying L'Hopital's Rule

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To find the limit of x^(2x) as x approaches 0, we can use logarithmic differentiation. Taking the natural logarithm of both sides, we get ln(y) = 2x * ln(x). Differentiating implicitly with respect to x, we have (1/y) * dy/dx = 2 * ln(x) + 2x * (1/x). Simplifying this expression, we get dy/dx = 2 * ln(x) + 2. Now, we substitute x = 0 into this expression to find the limit. However, ln(0) is undefined, so we cannot directly substitute x = 0. Instead, we can use L'Hôpital's rule. Taking the derivative of both the numerator and denominator, we get dy/dx = (2/x) / (1/x^2) = 2x^2/x = 2x. Now, we substitute x = 0 into this expression to find the limit. The limit of 2x as x approaches 0 is 0. Therefore, the limit of x^(2x) as x approaches 0 is 1.

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