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

Question

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

Christopher Agresta · Accepted Answer

#-1/4#

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#

Aurora Alvarado · Answer

Multiply by the conjugate of the numerator over itself.

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#

Isabella Ackerman · Answer

undefined To find the limit of ((1-x)^(1/4)-1)/x as x approaches 0, we can use algebraic manipulation and the limit definition.

First, let's simplify the expression by multiplying the numerator and denominator by the conjugate of the numerator, which is ((1-x)^(1/4)+1):

((1-x)^(1/4)-1)/x * ((1-x)^(1/4)+1)/((1-x)^(1/4)+1)

This simplifies to:

((1-x)^(1/4))^2 - 1^2 / x * ((1-x)^(1/4)+1)

Simplifying further:

(1-x)^(1/2) - 1 / x * ((1-x)^(1/4)+1)

Now, we can apply the limit definition. As x approaches 0, we substitute 0 into the expression:

(1-0)^(1/2) - 1 / 0 * ((1-0)^(1/4)+1)

Simplifying:

1^(1/2) - 1 / 0 * (1^(1/4)+1)

This further simplifies to:

1 - 1 / 0 * (1+1)

Since division by 0 is undefined, the limit does not exist.

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

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