How do you find #lim ((1-x)^(1/4)-1)/x# as #x->0# using l'Hospital's Rule?

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Answer 1 · 2017-04-14T18:45:04

#-1/4#

Take the derivative of the top and bottom. Don't forget chain rule:

#lim_(xrarr0) (1/4)(-1)(1-x)^(-3/4)/1#

Plug in #0# for #x#:

#(1/4)(-1)(1)^(-3/4)/1=-1/4#

Answer 2 · 2017-04-14T19:40:22

Alt approach:

#lim_(x to 0) ((1-x)^(1/4)-1)/x#

#= lim_(x to 0) ((1-1/4 x + O(x^2))-1)/x#

#= lim_(x to 0) -1/4 + O(x) = - 1/4#