How do I know what to substitute to simplify a limit: $\lim\limits_{x \rightarrow 1}{\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}$?

Question

I still do not know the answer, so I am looking for hints and tips here. Note that I want to do this with substitution, or at least not by taking derivatives.

Evaluate: (that's a cube root on the numerator)

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

The example in the book suggests substituting to make the problem simpler, but I tried using $t = \sqrt{x}-1$ and the numerator also, but I still get a divide by zero in either case.

How do you make an educated guess on what to use for $t$ and substitute?

Answer 1

Well, you want a substitution so that you can remove everything in terms of $x$ and convert everything to being in terms of $t$. I recommend $t=\sqrt[6]{x},$ personally, since both $\sqrt[3]{x}$ and $\sqrt{x}$ can be expressed as powers of $\sqrt[6]{x},$ for non-negative $x$.

Once you've done that, the difference of squares and difference of cubes formulas should come in handy.

Answer 2

My approach, if I were to do it by algebra alone, would be to use the two identities $u^2-1=(u-1)(u+1)$ and $u^3-1=(u-1)(u^2+u+1)$ with $u=\sqrt{x}$ and $u=\sqrt[3]{x}$, respectively. So you get $$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1},$$ from which the limit is easily computed.

Answer 3

Expanding on Cameron's answer a bit, we have:

$$\frac {t^2 - 1} {t^3 - 1} = \frac {(t -1)(t+1)} {(t-1)(t^2 + t + 1)},$$ so $$\lim_{t\to 1} \frac {t^2 - 1} {t^3 - 1} = 2/3.$$

Answer 4

$$ \lim_{x \to 1}{\sqrt[3]{x} - 1 \over \sqrt{x} - 1} = \lim_{x \to 1}{\left(1/3\right)\,x^{-2/3} \over \left(1/2\right)\,x^{-1/2}} = {2 \over 3} $$