How do you find the limit #((1-x)^(1/4)-1)/x# as #x->0#?

According to the binomial expansion we have

#(1-x)^(1/4)=1+1/4(-x)+(1/4(1/4-1))/(2!)(-x)^2+cdots# so

#((1-x)^(1/4)-1)/x=((1+1/4(-x)+(1/4(1/4-1))/(2!)(-x)^2+cdots)-1)/x#
#=-1/4+(1/4(1/4-1))/(2!)x+O(x^2)#

Here #O(x^2)# means an infinite sum of terms with at least order #2#

Then

#lim_(x->0)((1-x)^(1/4)-1)/x=-1/4#

Another approach is rationalizing

#((1-x)^(1/4)-1)/x=1/x((1-x)^(1/2)-1)/((1-x)^(1/4)+1)=1/x(1-x-1)/(((1-x)^(1/4)+1)((1-x)^(1/2)+1)) = -1/(((1-x)^(1/4)+1)((1-x)^(1/2)+1))# so

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

For positive integer #n#, we have

#u^n-v^n = (u-v)(u^(n-1)+u^(n-2)v+u^(n-3)v^2 + * * * +uv^(n-2)+v^(n-1))#.

So

#(a^(1/n)-b^(1/n))(a^((n-1)/n) + a^((n-2)/n)b^(1/n) +a^((n-3)/n)b^(2/n) + * * * +a^(1/n)b^((n-2)/n) +b^((n-1)/n)) = a-b #

#lim_(xrarr0) ((1-x)^(1/4)-1)/x = lim_(xrarr0)(((1-x)^(1/4)-1)/x)(((1-x)^(3/4)+(1-x)^(2/4)+(1-x)^(1/4)+1)/((1-x)^(3/4)+(1-x)^(2/4)+(1-x)^(1/4)+1))#

# = lim_(xrarr0) ((1-x)-1)/(x((1-x)^(3/4)+(1-x)^(2/4)+(1-x)^(1/4)+1))#

# = lim_(xrarr0) (-x)/(x((1-x)^(3/4)+(1-x)^(2/4)+(1-x)^(1/4)+1))#

# = lim_(xrarr0) (-1)/((1-x)^(3/4)+(1-x)^(2/4)+(1-x)^(1/4)+1)#

# = (-1)/((1-0)^(3/4)+(1-0)^(2/4)+(1-0)^(1/4)+1)#

# = (-1)/4#