Limit of (1 − x) tan(πx/2) at x = 1
Evaluate the limit: $$ L = \lim_{x \to 1} (1-x)\tan\frac{\pi x}{2}. $$
Explanation of Method
As $x\to1$, $\frac{\pi x}{2}\to\frac{\pi}{2}$, where $\tan$ has a vertical asymptote, while $1-x\to0$ — an indeterminate form $0\cdot\infty$. Rather than substituting $y=1-x$ directly, it is equally effective to measure how far the tangent's argument is from the pole at $\frac\pi2$: setting $t=\frac\pi2-\frac{\pi x}{2}$ makes $t\to0$ exactly when $x\to1$, and rewrites the tangent as a cotangent of $t$, which behaves like $\frac1t$ near $t=0$.
- Substitute $t=\frac\pi2-\frac{\pi x}2$, and express $1-x$ in terms of $t$.
- Rewrite $\tan\frac{\pi x}2$ as $\cot t$ using the co-function identity.
- Approximate $\tan t\approx t$ for small $t$ and take the limit.
Step 1 · Substitute $t=\frac\pi2-\frac{\pi x}{2}$
Let $t = \dfrac\pi2-\dfrac{\pi x}2$, so that $\dfrac{\pi x}2 = \dfrac\pi2-t$. As $x\to1$, $t\to0^+$ (for $x$ slightly less than $1$; the argument is symmetric for $x\to1^-$ or $1^+$). Solving for $1-x$ in terms of $t$: since $t=\dfrac\pi2(1-x)$, we get:
$$ 1-x = \frac{2t}{\pi}. $$
Step 2 · Rewrite the tangent as a cotangent
Using the co-function identity $\tan\left(\dfrac\pi2-t\right)=\cot t = \dfrac1{\tan t}$:
$$ \tan\frac{\pi x}{2} = \tan\left(\frac\pi2-t\right) = \cot t = \frac{1}{\tan t}. $$
So the original limit becomes:
$$ L = \lim_{t \to 0} \frac{2t}{\pi}\cdot\frac{1}{\tan t}. $$
Step 3 · Approximate $\tan t$ and take the limit
For small $t$, $\tan t \approx t$, so $\dfrac{1}{\tan t}\approx\dfrac1t$. Substituting:
$$ L = \lim_{t\to0}\frac{2t}{\pi}\cdot\frac1t = \lim_{t\to0}\frac{2}{\pi} = \frac2\pi. $$
Summary Table
| Quantity | In terms of $t=\frac\pi2-\frac{\pi x}2$ | Behavior as $t\to0$ |
|---|---|---|
| $1-x$ | $\dfrac{2t}{\pi}$ | $\to0$ |
| $\tan\dfrac{\pi x}2$ | $\cot t \approx \dfrac1t$ | $\to\infty$ |
| Product | $\dfrac{2t}{\pi}\cdot\dfrac1t$ | $\to\dfrac2\pi$ |
Additional Notes
- This is the same limit that appears elsewhere on this site solved with the substitution $y=1-x$ directly (leading to the same co-function identity $\tan\left(\frac\pi2-\frac{\pi y}2\right)=\cot\frac{\pi y}2$) — the version here instead measures the "distance from the pole" of the tangent function, which is a slightly more direct way to see why the pole and the vanishing factor exactly balance.
- Whichever substitution is used, the underlying mechanism is the same: a linear factor vanishing at exactly the rate needed to cancel the blow-up of $\cot$ (or $\tan$) near its pole, leaving a finite ratio determined only by the coefficients in the two linear arguments.
- For a fully rigorous version (avoiding the approximation $\tan t\approx t$), one can use the exact identity $t\cot t = \dfrac{t}{\sin t}\cos t \to 1\cdot1=1$ as $t\to0$, using the standard limit $\lim_{t\to0}\frac{t}{\sin t}=1$ together with continuity of $\cos$ at $0$.
Result
The value of the limit is: