Limit of a square root minus a cube root at 0
Evaluate the limit: $$ L = \lim_{x \to 0} \left(\sqrt{1-2x}-\sqrt[3]{1-3x}\right). $$
Explanation of Method
Both $\sqrt{1-2x}$ and $\sqrt[3]{1-3x}$ are continuous at $x=0$ and both equal $1$ there, so their difference tends to $0-0=0$ by continuity alone — no indeterminate form is really present. What makes this problem interesting is that expanding each term to first order in $x$ shows the linear terms cancel exactly, so to see the true rate at which the difference vanishes, the binomial expansion must be carried out to second order.
- Expand $\sqrt{1-2x}$ using the binomial approximation to second order.
- Expand $\sqrt[3]{1-3x}$ to second order as well.
- Subtract the two expansions and observe that the linear terms cancel.
- Take the limit, using the surviving quadratic term.
Step 1 · Expand $\sqrt{1-2x}$ to second order
Using the binomial expansion $(1+z)^{1/2}=1+\dfrac z2-\dfrac{z^2}8+O(z^3)$ with $z=-2x$:
$$ \sqrt{1-2x} = 1+\frac{-2x}{2}-\frac{(-2x)^2}{8}+O(x^3) = 1-x-\frac{x^2}{2}+O(x^3). $$
Step 2 · Expand $\sqrt[3]{1-3x}$ to second order
Using the binomial expansion $(1+z)^{1/3}=1+\dfrac z3-\dfrac{z^2}9+O(z^3)$ with $z=-3x$:
$$ \sqrt[3]{1-3x} = 1+\frac{-3x}{3}-\frac{(-3x)^2}{9}+O(x^3) = 1-x-x^2+O(x^3). $$
Step 3 · Subtract the two expansions
Subtracting:
$$ \sqrt{1-2x}-\sqrt[3]{1-3x} = \left(1-x-\frac{x^2}2\right) - \left(1-x-x^2\right) + O(x^3) = -\frac{x^2}2+x^2+O(x^3) = \frac{x^2}{2}+O(x^3). $$
Notice that both the constant term ($1-1=0$) and the linear term ($-x-(-x)=0$) cancel exactly, leaving a surviving quadratic term $\frac{x^2}2$.
Step 4 · Take the limit
Since $\dfrac{x^2}2+O(x^3)\to0$ as $x\to0$:
$$ L = \lim_{x\to0}\left(\sqrt{1-2x}-\sqrt[3]{1-3x}\right) = \lim_{x\to0}\left(\frac{x^2}{2}+O(x^3)\right) = 0. $$
Summary Table
| Expansion | To second order |
|---|---|
| $\sqrt{1-2x}$ | $1-x-\dfrac{x^2}2+O(x^3)$ |
| $\sqrt[3]{1-3x}$ | $1-x-x^2+O(x^3)$ |
| Difference | $\dfrac{x^2}2+O(x^3) \ \to \ 0$ |
Additional Notes
- Both roots share the exact same linear approximation, $1-x$, near $x=0$ — this is because $z=-2x$ divided by $2$ (for the square root) and $z=-3x$ divided by $3$ (for the cube root) both simplify to the same coefficient $-x$; the difference between the two roots only becomes visible at the next order, where the quadratic coefficients ($-\frac12$ versus $-1$) differ.
- Expanding only to first order (as is often "good enough" to establish that a limit is $0$) would have concluded correctly here that the limit is $0$, but would have missed the fact that the difference vanishes specifically like $\frac{x^2}2$ — a distinction that matters if this expression were itself divided by, say, $x^2$ in a larger problem.
- This kind of exact cancellation of lower-order terms, revealing a higher-order rate, is common whenever two different functions share the same first-order (tangent line) behavior at a point but differ in their curvature (second-order behavior).
Result
The value of the limit is: