Question

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

Answer 1

`-1/4`

Explanation:

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`

Answer 2

Multiply by the conjugate of the numerator over itself.

Explanation:

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`