Limit of (arcsin x / x) to the power 1/x²
Evaluate the limit: $$ L = \lim_{x \to 0} \left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}. $$
Explanation of Method
As $x\to0$, $\dfrac{\arcsin x}x\to1$ while the exponent $\dfrac1{x^2}\to\infty$, giving the indeterminate form $1^\infty$. The Maclaurin series for $\arcsin x$ shows that the base is $1$ plus a term of order $x^2$; raising an expression of the form $1+cx^2$ to the power $\frac1{x^2}$ is the standard setup that converges to $e^{c}$, by the defining property of the exponential function.
- Expand $\arcsin x$ as a Maclaurin series and simplify $\frac{\arcsin x}{x}$.
- Identify the coefficient $c$ in the resulting form $1+cx^2+O(x^4)$.
- Apply the standard limit $(1+cx^2)^{1/x^2}\to e^c$ as $x\to0$.
Step 1 · Expand $\arcsin x$ and simplify the base
The Maclaurin series for $\arcsin x$ is:
$$ \arcsin x = x + \frac{x^3}{6} + O(x^5). $$
Dividing by $x$:
$$ \frac{\arcsin x}{x} = 1 + \frac{x^2}{6} + O(x^4). $$
Step 2 · Identify the coefficient of $x^2$
The base has the form $1+cx^2+O(x^4)$ with $c=\dfrac16$.
Step 3 · Apply the standard exponential limit
Recall the defining property of $e$: for any fixed constant $c$,
$$ \left(1+cx^2+O(x^4)\right)^{1/x^2} \longrightarrow e^{c} \qquad \text{as } x \to 0. $$
(This follows by taking $\ln$, using $\ln(1+cx^2+O(x^4))\approx cx^2$, dividing by $x^2$ to get $c$, and exponentiating.) With $c=\dfrac16$:
$$ L = \left(1+\frac{x^2}{6}+O(x^4)\right)^{1/x^2} \longrightarrow e^{1/6}. $$
Summary Table
| Quantity | Series / value |
|---|---|
| $\arcsin x$ | $x+\dfrac{x^3}{6}+O(x^5)$ |
| $\dfrac{\arcsin x}{x}$ | $1+\dfrac{x^2}{6}+O(x^4)$ |
| $L$ | $e^{1/6}$ |
Additional Notes
- This limit mirrors the earlier one involving $\arctan x$: there, $\frac{\arctan x}{x}=1-\frac{x^2}3+O(x^4)$ gave $\left(\frac{\arctan x}x\right)^{1/x^2}\to e^{-1/3}$; here, the opposite sign in the Maclaurin series of $\arcsin$ (which has $+\frac{x^3}6$ instead of $-\frac{x^3}3$) gives the positive exponent $e^{1/6}$ instead.
- Taking logarithms is the standard first move for any $1^\infty$ indeterminate form, since it converts a power into a product; the shortcut used here (recognizing the form $(1+cx^2)^{1/x^2}\to e^c$ directly) is simply a compressed version of that same log-and-exponentiate procedure.
- The general pattern is: for any function $g(x)\to0$ with $g(x)\sim cx^2$ as $x\to0$, $(1+g(x))^{1/x^2}\to e^c$ — only the coefficient of the $x^2$ term in $g$'s expansion determines the final answer.
Result
The value of the limit is: