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

Question

10869 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-30T20:11:05

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#