Evaluating Limit with Radical: Can You Help?

[Precursor]

Evaluate if the following limit exists:

$$lim_{x\rightarrow1}\frac{\sqrt[3]{x}-1}{x-1}$$


My work:

$$lim_{x\rightarrow1}\frac{\sqrt[3]{x^{2}}-1}{(x-1)(\sqrt[3]{x}+1)}}$$

I did the conjugate, but I'm still left with a radical in the numerator, and I can't seem to factor any further. Can someone help me out?

 

[Mark44]

You have the right idea, but haven't quite carried it off. The radical term in your numerator is x^(2/3) != x^(1/3) in the original limit expression.

x - 1 can be thought of as the difference of cubes, as $(\sqrt[3]{x})^3 - 1^3$.

A difference of cubes can be factored like so:
$$(a^3 - b^3) = (a - b)(a^2 + ab + b^2)$$.

 

[Precursor]

Thanks, that helped for that question. But how about a question like the following(we have not yet been taught the difference of fourth root):

$$lim_{x\rightarrow0}\frac{\sqrt[4]{1+x}-1}{x}$$

Could I do a difference of squares and then the conjugate?

 

[Mark44]

How about if you multiplied by 1 (which is always legal)?

The 1 I am thinking of looks like
$$\frac{(u + 1)(u^2 + 1)}{(u + 1)(u^2 + 1)}$$

where $$u = \sqrt[4]{1 + x}$$.

The idea is that you have something that looks like u - 1 in the numerator, and I want to turn it into u^4 - 1. The other factors to make this happen are (u + 1) and (u^2 + 1).

No guarantees that this will work, but it might make it so that some things cancel so that you don't get zero in both the numerator and denominator.