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

Question

Evelyn Agard · Accepted Answer

#1# #L = lim_{x o 0} x^{2x}# consider #y = x^{2x}# so that # ln y = 2x ln x# # ln L = lim_{x o 0} 2x ln x# # = 2 lim_{x o 0} (ln x)/(1/x)# so it is now indeterminate Applying L'Hopital's Rule #ln L = 2 lim_{x o 0} (ln x)/(1/x) = 2 lim_{x o 0} (1/x)/(-1/x^2) = -2 lim_{x o 0} x = 0# so # lim_{x o 0} ln L = 0 \implies lim_{x o 0} L = 1#

Scarlett Allison · Answer

undefined 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.