How do you find the limit of #x^sqrtx# as x approaches 0?

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Answer 1 · 2016-07-04T08:02:25

1

#lim_{x to 0} x^{sqrt x}#

#= lim_{x to 0} e^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))#

#lim_{x to 0} ln x/ (1/sqrt x) = oo/oo# ie indeterminate, so we can use L'Hopital

#= 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 x^(1/2))#

#= e^0 = 1#