How to evaluate x limits with fractional powers

Expert intro: This limit looks awkward at first because the numerator and denominator both go to 0. The useful move is to rewrite the fractional exponents so they share the same base power.

Detailed walkthrough

Let

$L=\lim_{x\to 1}\frac{x^{\frac16}-1}{x^{\frac13}-1}.$

Notice that

$x^{\frac13}=\left(x^{\frac16}\right)^2.$

So if we set

$t=x^{\frac16},$

then as $x\to 1$, we also have $t\to 1$, and the limit becomes

$L=\lim_{t\to 1}\frac{t-1}{t^2-1}.$

Factor the denominator:

$t^2-1=(t-1)(t+1).$

So

$L=\lim_{t\to 1}\frac{t-1}{(t-1)(t+1)}=\lim_{t\to 1}\frac{1}{t+1}.$

Now substitute $t=1$:

$L=\frac{1}{2}.$

So the value of the limit is

$\boxed{\frac12}.$

💡 Pitfall guide

A common slip is to try canceling $x-1$ directly, but these are not linear expressions. Another mistake is treating $x^{1/6}$ and $x^{1/3}$ as unrelated; once you notice $x^{1/3}=(x^{1/6})^2$, the algebra gets much cleaner.

🔄 Real-world variant

If the denominator were $x^{\frac23}-1$ instead, you could set $t=x^{1/6}$ again and rewrite it as $t^4-1$. Then factoring would give a different constant limit. The same idea works whenever the exponents are multiples of a common fraction.

🔍 Related terms

fractional exponents, indeterminate form, factorization