Constant that makes a logarithm power continuous

Let $$ f(x) = \ln\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}, \quad x \neq 0, \qquad f(0) = \alpha. $$ Determine the value of $\alpha$ for which $f$ is continuous at $x=0$.

Step 1 · Simplify $f(x)$

Using $\ln(a^b) = b\ln a$:

$$ f(x) = \ln\left(\frac{\sin x}{x}\right)^{1/x^2} = \frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right). $$

Step 2 · Expand $\dfrac{\sin x}{x}$ near $x=0$

The Taylor series for $\sin x$ near $0$ is $\sin x = x - \dfrac{x^3}{6} + O(x^5)$, so:

$$ \frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4). $$

Step 3 · Approximate the logarithm

Since $\dfrac{\sin x}{x} \to 1$ as $x\to0$, use $\ln(1+u)\approx u$ for small $u = -\dfrac{x^2}{6}+O(x^4)$:

$$ \ln\left(\frac{\sin x}{x}\right) = \ln\left(1-\frac{x^2}{6}+O(x^4)\right) \approx -\frac{x^2}{6} + O(x^4). $$

Step 4 · Divide by $x^2$ and take the limit

Substituting into $f(x)$:

$$ f(x) = \frac{1}{x^2}\left(-\frac{x^2}{6}+O(x^4)\right) = -\frac16 + O(x^2). $$

Taking the limit as $x \to 0$:

$$ \lim_{x\to0} f(x) = -\frac16. $$

For $f$ to be continuous at $x=0$, we need $\alpha = f(0) = \lim_{x\to0}f(x)$, so:

$$ \alpha = -\frac16. $$

Summary Table

Additional Notes

• This limit is equivalent to saying $\left(\dfrac{\sin x}{x}\right)^{1/x^2} \to e^{-1/6}$ as $x\to0$ — exponentiating $\alpha=-\frac16$ recovers the base of that related $1^\infty$-type limit.
• Only the $x^2$-order term of the Taylor expansion of $\sin x$ mattered here, since higher-order corrections vanish once divided by $x^2$ and the limit is taken — this is a common pattern whenever a logarithm of a ratio is divided by a matching power.
• The constant $\frac16$ is directly inherited from the coefficient $\frac16$ in the Taylor series $\sin x = x - \frac{x^3}{6}+\cdots$; a different oscillatory function in place of $\sin x$ would produce a different constant here.