Limit of (arctan x / x) to the power 1/x²
Evaluate the limit: $$ L = \lim_{x \to 0} \left( \frac{\arctan x}{x} \right)^{\frac{1}{x^2}}. $$
Explanation of Method
As $x \to 0$, $\dfrac{\arctan x}{x} \to 1$ while the exponent $\dfrac{1}{x^2} \to \infty$, giving the indeterminate form $1^\infty$. The standard technique is to take the natural logarithm of the expression, turning the power into a product that can be evaluated with a Taylor (Maclaurin) series expansion, and then exponentiate the result at the end.
- Expand $\arctan x$ as a Maclaurin series and simplify $\dfrac{\arctan x}{x}$.
- Take the logarithm of $L$ and use $\ln(1+u) \approx u$ for small $u$.
- Evaluate the resulting limit for $\ln L$.
- Exponentiate to recover $L$.
Step 1 · Expand $\arctan x$ and simplify the base
The Maclaurin series for $\arctan x$ is:
$$ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots = x - \frac{x^3}{3} + O(x^5). $$
Dividing by $x$ gives the base of the limit:
$$ \frac{\arctan x}{x} = 1 - \frac{x^2}{3} + O(x^4). $$
Step 2 · Take the logarithm of $L$
Let $L = \displaystyle\lim_{x\to 0}\left(\frac{\arctan x}{x}\right)^{1/x^2}$. Taking $\ln$ of both sides and using $\ln(a^b) = b\ln a$:
$$ \ln L = \lim_{x \to 0} \frac{1}{x^2} \ln\left( 1 - \frac{x^2}{3} + O(x^4) \right). $$
Since $-\dfrac{x^2}{3} + O(x^4) \to 0$ as $x \to 0$, we can use the small-argument approximation $\ln(1+u) \approx u$ for $u \to 0$:
$$ \ln\left( 1 - \frac{x^2}{3} + O(x^4) \right) \approx -\frac{x^2}{3} + O(x^4). $$
Step 3 · Evaluate the limit for $\ln L$
Substituting this approximation:
$$ \ln L = \lim_{x \to 0} \frac{1}{x^2}\left( -\frac{x^2}{3} + O(x^4) \right) = \lim_{x \to 0} \left( -\frac{1}{3} + O(x^2) \right) = -\frac{1}{3}. $$
Step 4 · Exponentiate to find $L$
Since $\ln L = -\dfrac{1}{3}$, exponentiating both sides gives:
$$ L = e^{-\frac{1}{3}}. $$
Summary Table
| Quantity | Series / approximation | Limit as $x \to 0$ |
|---|---|---|
| $\arctan x$ | $x - \dfrac{x^3}{3} + O(x^5)$ | — |
| $\dfrac{\arctan x}{x}$ | $1 - \dfrac{x^2}{3} + O(x^4)$ | $1$ |
| $\ln L$ | $-\dfrac{1}{3} + O(x^2)$ | $-\dfrac{1}{3}$ |
Additional Notes
- Taking the logarithm is the standard trick for any $1^\infty$, $0^0$, or $\infty^0$ indeterminate form, since it converts a power into a product.
- Only the lowest-order nonzero term of the series matters here, since it is divided by $x^2$ and all higher-order terms vanish as $x \to 0$.
- The same approach generalizes: for $\left(1 + c\,x^2 + O(x^4)\right)^{1/x^2}$, the limit is always $e^{c}$.
Result
The value of the limit is: