Evaluate the limit of a cube root expression at zero

Key concept: This limit is one of those cases where the expression looks awkward, but a clean algebraic or derivative idea makes it simple. The numerator is set up to behave like a difference quotient.

Step by step

We want

$\lim_{x\to 0}\frac{(3x+1)^{\frac13}-1}{x}$

Notice that this matches the derivative of

$g(x)=x^{1/3}$

at the point $x=1$, because

$\frac{(3x+1)^{1/3}-1}{x} = 3\cdot \frac{(3x+1)^{1/3}-1^{1/3}}{3x}$

Let

$u=3x+1,$ so as $x\to 0$, we have $u\to 1$ and $du=3\,dx$.

Then

=3\lim_{u\to 1}\frac{u^{1/3}-1}{u-1}$$ Now use the derivative of $u^{1/3}$ at $u=1$: $$\frac{d}{du}(u^{1/3})=\frac13u^{-2/3}$$ So at $u=1$, $$\lim_{u\to 1}\frac{u^{1/3}-1}{u-1}=\frac13$$ Therefore, $$3\cdot\frac13=1$$ ### Final answer $$\boxed{1}$$ ### Pitfall alert A common mistake is to plug in $x=0$ directly and conclude the fraction is undefined, so the limit does not exist. The numerator and denominator both go to zero, which is exactly when a limit often still exists. Another trap is forgetting the factor of 3 from the inner function $3x+1$. ### Try different conditions If the numerator were $(ax+1)^{1/3}-1$, the same idea would give $$\lim_{x\to 0}\frac{(ax+1)^{1/3}-1}{x}=\frac{a}{3}$$ provided the limit is taken near $x=0$ and the cube root branch is the usual real one. ### Further reading difference quotient, chain rule, cube root limit