Find lim as x to pi (pi-x) tan x/2

Find $\displaystyle\lim_{x\to\pi}\bigl(\pi-x\bigr)\tan\frac{x}{2}$.

Step 1: Identify the indeterminate form

As $x\to\pi$ the factor $\pi-x$ tends to $0$, while $\tan\dfrac{x}{2}$ becomes infinite: it runs to $+\infty$ as $x\to\pi^{-}$ and to $-\infty$ as $x\to\pi^{+}$. In either case the product is of the form $0\cdot\infty$.

Step 2: Substitute $t = \pi-x$

Put $t = \pi-x$, so that $x = \pi-t$ and $t\to0$. Then

\[ \frac{x}{2} = \frac{\pi-t}{2} = \frac{\pi}{2}-\frac{t}{2}, \qquad \tan\frac{x}{2} = \tan\left(\frac{\pi}{2}-\frac{t}{2}\right) = \cot\frac{t}{2}, \]

and the expression becomes

\[ \bigl(\pi-x\bigr)\tan\frac{x}{2} = t\cot\frac{t}{2} = \frac{t}{\tan\dfrac{t}{2}}. \tag{1} \]

Step 3: Evaluate by the standard equivalence

Since $\tan u\sim u$ as $u\to0$, with $u = \dfrac{t}{2}$: $\frac{t}{\tan\dfrac{t}{2}} \sim \frac{t}{\dfrac{t}{2}} = 2$.

Step 4: The same by l'Hôpital's rule

Expression (1) is of the form $\dfrac00$, and $\dfrac{d}{dt}\tan\dfrac{t}{2} = \dfrac{1}{2\cos^{2}\dfrac{t}{2}}$, so

\[ \lim_{t\to0}\frac{t}{\tan\dfrac{t}{2}} = \lim_{t\to0}\frac{1}{\dfrac{1}{2\cos^{2}\dfrac{t}{2}}} = \lim_{t\to0}2\cos^{2}\frac{t}{2} = 2\cos^{2}0 = 2. \]

Step 5: Verification

\[x = \pi-0.1:\ 1.9983331; \qquad x = \pi+0.1:\ 1.9983331 \;\;\checkmark\]

\[x = \pi\pm0.01:\ 1.9999833 \;\;\checkmark\]

\[\text{Both signs of } \tan\tfrac{x}{2} \text{ are matched by the sign of } \pi-x,\ \text{so the limit is two-sided} \;\;\checkmark\]