Limit of the n-th root of 1+x, minus 1
Let $n$ be a positive integer. Evaluate the limit: $$ L = \lim_{x \to 0} \left(\sqrt[n]{1+x}-1\right). $$
Explanation of Method
Since $\sqrt[n]{1+x}=(1+x)^{1/n}$ is continuous at $x=0$ for every positive integer $n$, this limit could be found immediately by direct substitution. The binomial approximation makes the underlying rate of vanishing explicit: expanding $(1+x)^{1/n}$ to first order in $x$ shows that the whole expression behaves like $\frac xn$ near $x=0$, which is a useful fact in its own right when this expression appears as a building block inside larger limits.
- Expand $(1+x)^{1/n}$ using the binomial approximation for small $x$.
- Simplify $\sqrt[n]{1+x}-1$ using that expansion.
- Take the limit as $x\to0$.
Step 1 · Expand $(1+x)^{1/n}$ for small $x$
The binomial approximation for $(1+x)^p$ near $x=0$ (valid for any real exponent $p$) is:
$$ (1+x)^p = 1+px+O(x^2). $$
With $p=\dfrac1n$:
$$ (1+x)^{1/n} = 1+\frac{x}{n}+O(x^2). $$
Step 2 · Simplify the expression
Subtracting $1$:
$$ \sqrt[n]{1+x}-1 = \left(1+\frac xn+O(x^2)\right)-1 = \frac xn + O(x^2). $$
Step 3 · Take the limit
As $x\to0$, the term $\frac xn\to0$ and the higher-order correction $O(x^2)\to0$ as well, so:
$$ L = \lim_{x\to0}\left(\sqrt[n]{1+x}-1\right) = \lim_{x\to0}\left(\frac xn+O(x^2)\right) = 0. $$
Summary Table
| Quantity | Expansion for small $x$ | Limit as $x\to0$ |
|---|---|---|
| $(1+x)^{1/n}$ | $1+\dfrac xn+O(x^2)$ | $1$ |
| $\sqrt[n]{1+x}-1$ | $\dfrac xn+O(x^2)$ | $0$ |
Additional Notes
- The value of the limit itself ($0$) does not depend on $n$ at all, since $\sqrt[n]{1+x}\to\sqrt[n]{1}=1$ regardless of which positive integer $n$ is chosen — but the rate at which the expression approaches $0$ does depend on $n$: it behaves like $\frac xn$, so larger $n$ makes the expression vanish more slowly (relative to $x$) as $x\to0$.
- This rate, $\sqrt[n]{1+x}-1\sim\frac xn$, is one of the most frequently used building blocks in more elaborate limits, exactly analogous to $\ln(1+x)\sim x$ and $e^x-1\sim x$ — all three are first-order approximations near $x=0$ that appear repeatedly when simplifying compound expressions.
- For $n=1$, the expression is trivially $x$ itself (no approximation needed), and the rate $\frac x1=x$ correctly recovers that special case.
Result
The value of the limit is: