Find the limit, as x approaches 0, of the quantity (sin x)/x raised to the power 1/(cos x − 1).

Step-by-step solution

1. 1

   Classify the indeterminate form. As $x \to 0$ the base $\frac{\sin x}{x} \to 1$ while the exponent $\frac{1}{\cos x - 1} \to \pm\infty$ (it is negative, since $\cos x - 1 < 0$). That is the $1^{\infty}$ form, whose value is genuinely undetermined — it can be anything — so it must be resolved, not guessed.

2. 2

   Take logarithms to turn the power into a quotient. Let $L$ be the limit. Then

   $\ln L = \lim_{x \to 0} \frac{\ln\!\left(\frac{\sin x}{x}\right)}{\cos x - 1}$

   This is now a plain $0/0$ limit, which series expansions handle cleanly.

3. 3

   Expand the base to order x². From $\sin x = x - \frac{x^3}{6} + O(x^5)$:

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

   Then, using $\ln(1 + u) = u + O(u^2)$ with $u = -\frac{x^2}{6}$:

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

   The leading term is quadratic, not linear — that is why the denominator also has to be taken to order $x^2$.

4. 4

   Expand the denominator to the same order. From $\cos x = 1 - \frac{x^2}{2} + O(x^4)$:

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

5. 5

   Divide and exponentiate. The $x^2$ factors cancel and the signs cancel too:

   $\ln L = \lim_{x \to 0} \frac{-x^2/6}{-x^2/2} = \frac{1/6}{1/2} = \frac{1}{3}$

   Therefore

   $L = e^{1/3} \approx 1.39561$

6. 6

   Confirm numerically. Evaluating the original expression at $x = 10^{-2}$ gives $1.3956179$ and at $x = 10^{-4}$ gives $1.3956124$, against $e^{1/3} = 1.3956124$. The function is even, so $x < 0$ gives the same values — the two-sided limit exists.