Find the limit of (tan x / x)^(1/x) as x approaches 0.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $x\to0$, $\tan x\sim x$ so the base $\tfrac{\tan x}{x}\to1$, while the exponent $\tfrac1x\to\pm\infty$. That is $1^{\infty}$ - genuinely indeterminate, because a base slightly above $1$ raised to a huge power can go anywhere.

2. 2

   Take logarithms to move the exponent down. Setting $L$ for the limit,

   $\ln L=\lim_{x\to0}\frac{1}{x}\ln\!\left(\frac{\tan x}{x}\right)$

   The whole question is now: how fast does the base approach $1$ compared with how fast the exponent blows up?

3. 3

   Expand the base to enough orders. From $\tan x=x+\tfrac{x^3}{3}+O(x^5)$,

   $\frac{\tan x}{x}=1+\frac{x^2}{3}+O(x^4)$

   The key number is the $x^2$: the base approaches $1$ quadratically, faster than the exponent's single power of $x$ blows up.

4. 4

   Apply $\ln(1+u)\approx u$. With $u=\tfrac{x^2}{3}+O(x^4)$,

   $\ln\!\left(\frac{\tan x}{x}\right)=\frac{x^2}{3}+O(x^4)\;\Longrightarrow\;\frac{1}{x}\ln\!\left(\frac{\tan x}{x}\right)=\frac{x}{3}+O(x^3)$

5. 5

   Take the limit and undo the logarithm. Since $\tfrac{x}{3}\to0$ from both sides, $\ln L=0$ and

   $L=e^{0}=1$

6. 6

   Confirm numerically from both sides. At $x=0.001$ the expression equals $1.000333$; at $x=-0.001$ it equals $0.999667$. Both squeeze toward $1$ as $x\to0$, and the two-sided agreement confirms the limit exists ✓.