How do you find the limit of #x^sqrtx# as x approaches 0 using l'hospital's rule?

Question

Charles Anctil · Accepted Answer

1

start by using log properties to get that exponent down so # lim_{x to 0} x^sqrt(x)# # = lim_{x to 0} exp( ln x^sqrt(x) )# # = exp( lim_{x to 0} ln x^sqrt(x) )# # = exp (lim_{x to 0} sqrt(x) ln x )# #=exp ( lim_{x to 0} ln x/ (1/ sqrt(x))) # Now # lim_{x to 0} (ln x/ (1/ sqrt(x))) = - oo/oo# --> indeterminate so we use L'Hopital on that : # exp( lim_{x to 0} ln x/ ((x)^{-1/2})) = exp(lim_{x to 0} (1/x)/ (-1/2 x^(-3/2)) )# #= exp ( lim_{x to 0} (-2 x^{3/2}/x)) = exp ( lim_{x to 0} -2 sqrt(x)) # # = e^0# # = 1#

Aurora Alvarado · Answer

undefined To find the limit of x^sqrt(x) as x approaches 0 using L'Hospital's rule, we can rewrite the expression as e^(sqrt(x) * ln(x)). Taking the natural logarithm of both sides, we get ln(x^sqrt(x)) = sqrt(x) * ln(x). Now, we can apply L'Hospital's rule by differentiating the numerator and denominator separately. Differentiating sqrt(x) gives us 1/(2*sqrt(x)), and differentiating ln(x) gives us 1/x. Taking the limit as x approaches 0, we have (1/(2*sqrt(x))) / (1/x). Simplifying this expression, we get (x/(2*sqrt(x))) * (x/1). Canceling out x, we are left with 1/(2*sqrt(x)). Taking the limit as x approaches 0, we find that the limit of x^sqrt(x) as x approaches 0 is 0.