How to Evaluate Limits Involving Cube Roots Without L’Hôpital’s Rule

Limits with cube roots often produce a quotient that looks like $0/0$ when direct substitution is attempted. You can usually simplify it without L’Hôpital’s rule or a series by using the difference-of-cubes identity. The method works for real cube roots, including negative values.

The key identity

For any real numbers $u$ and $v$,

$$u^3-v^3=(u-v)(u^2+uv+v^2).$$

So, whenever $u$ and $v$ are not both zero,

$$u-v=\frac{u^3-v^3}{u^2+uv+v^2}.$$

When the roots in your limit are $u=\sqrt[3]{A(x)}$ and $v=\sqrt[3]{B(x)}$, their cubes are simply $A(x)$ and $B(x)$. This replaces a difference of roots with a difference of the expressions inside the roots. If that difference has a factor matching the original denominator, you can cancel it.

The quadratic factor $u^2+uv+v^2$ is nonnegative for real $u,v$ and equals zero only when both are zero. Thus the quotient identity is valid wherever its denominator is nonzero.

Worked example: a cube-root difference quotient

Evaluate

$$\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}.$$

Substitution gives $0/0$, so use the identity with $u=\sqrt[3]{1+x}$ and $v=1$:

$$\begin{aligned} \frac{\sqrt[3]{1+x}-1}{x} &=\frac{(\sqrt[3]{1+x})^3-1^3}{x\left((\sqrt[3]{1+x})^2+\sqrt[3]{1+x}+1\right)}\\ &=\frac{(1+x)-1}{x\left((\sqrt[3]{1+x})^2+\sqrt[3]{1+x}+1\right)}\\ &=\frac{1}{(\sqrt[3]{1+x})^2+\sqrt[3]{1+x}+1},\qquad x\ne 0. \end{aligned}$$

The simplified expression is continuous at $x=0$, so substitute $0$:

$$\lim_{x\to 0}\frac{1}{(\sqrt[3]{1+x})^2+\sqrt[3]{1+x}+1}=\frac{1}{1+1+1}=\frac13.$$

The original quotient is undefined at $x=0$, but a limit depends on values near $0$, not on the value at $0$. The cancellation is therefore legitimate for the limit.

A general version

Suppose $g$ is continuous near $a$, and consider

$$\lim_{x\to a}\frac{\sqrt[3]{g(x)}-\sqrt[3]{g(a)}}{x-a}.$$

Applying the identity gives, for $x\ne a$ where the quadratic factor is nonzero (equivalently, where $g(x)$ and $g(a)$ are not both zero),

$$\frac{g(x)-g(a)}{(x-a)\left((\sqrt[3]{g(x)})^2+\sqrt[3]{g(x)}\sqrt[3]{g(a)}+(\sqrt[3]{g(a)})^2\right)}.$$

This is useful when $g(x)-g(a)$ contains a factor of $x-a$. For example, take $g(x)=x$ and $a=8$:

$$\begin{aligned} \lim_{x\to 8}\frac{\sqrt[3]{x}-2}{x-8} &=\lim_{x\to 8}\frac{1}{(\sqrt[3]{x})^2+2\sqrt[3]{x}+4}\\ &=\frac{1}{4+4+4}=\frac{1}{12}. \end{aligned}$$

Here the numerator after applying the identity is $x-8$, which cancels the denominator. The remaining expression can be evaluated by continuity.

If $g$ is differentiable at $a$ and $g(a)\ne 0$, the same simplification also shows that the limit is

$$\frac{g'(a)}{3(\sqrt[3]{g(a)})^2}.$$

This last formula is optional; the algebraic identity itself is enough for many specific limits. When $g(a)=0$, the quadratic factor in the denominator tends to zero, so you must simplify further or analyze the limit directly rather than use this formula.

Common mistakes

• Using the square-root conjugate. Multiplying by $u+v$ does not eliminate a difference of cube roots. The relevant factor is $u^2+uv+v^2$.
• Dropping a term or sign. The correct identity is $u^3-v^3=(u-v)(u^2+uv+v^2)$. Check the result by expanding it if needed.
• Substituting before simplifying. If direct substitution gives $0/0$, that does not determine the limit. First expose and cancel the common factor.
• Canceling at a point without checking nearby values. Cancellation is valid for the limit when the factor is nonzero in a punctured neighborhood of the point. The expressions do not have to be defined at the point itself.
• Assuming every such limit is finite. If the remaining denominator tends to zero, or the numerator does not cancel the vanishing factor, the limit may be infinite or may fail to exist. Continue simplifying and check the one-sided behavior when necessary.

For a specific expression, identify the two cube roots, apply the identity, and then inspect whether the difference of their cubes factors by the denominator of the original quotient.