Evaluate the limit of cube root x minus 3 over x minus 27

Expert intro: This limit looks indeterminate at first glance, but it collapses neatly once you recognize the derivative pattern hiding in it.

Detailed walkthrough

We need to compute

$\lim_{x\to 27}\frac{\sqrt[3]{x}-3}{x-27}.$

Step 1: Recognize the structure

Let

$f(x)=x^{1/3}.$

Then the expression is

$\frac{f(x)-f(27)}{x-27},$

which is exactly the difference quotient for $f'(27)$.

Step 2: Differentiate

$f'(x)=\frac{1}{3}x^{-2/3}.$

So

$f'(27)=\frac{1}{3}\cdot 27^{-2/3}.$

Since

$27^{1/3}=3 \quad\Rightarrow\quad 27^{2/3}=9,$

we get

$f'(27)=\frac{1}{3\cdot 9}=\frac{1}{27}.$

Step 3: Final answer

$\boxed{\frac{1}{27}}.$

💡 Pitfall guide

The most common mistake is to plug in $x=27$ directly and call it $0/0$ without seeing the derivative form. Another frequent slip is differentiating $\sqrt[3]{x}$ as if it were $\frac{1}{3}x^{2/3}$ instead of $\frac{1}{3}x^{-2/3}$. The exponent must go negative.

🔄 Real-world variant

If the denominator were $x-a$ and the numerator were $\sqrt[3]{x}-\sqrt[3]{a}$, the same idea gives

$\lim_{x\to a}\frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a}=\frac{1}{3a^{2/3}}$

for $a>0$. Here $a=27$, which is why the value becomes $1/27$.

🔍 Related terms

difference quotient, derivative definition, cube root function