# Question #d7b66

Mar 4, 2017

${\lim}_{x \to 1} \left(1 - x\right) \tan \left(\frac{\pi x}{2}\right) = \frac{2}{\pi}$

#### Explanation:

The limit:

${\lim}_{x \to 1} \left(1 - x\right) \tan \left(\frac{\pi x}{2}\right)$

is in the indeterminate form $0 \cdot \infty$. We can then write it as:

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}}$

which is in the form $\frac{\infty}{\infty}$ so we can use l'Hospital's rule:

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = {\lim}_{x \to 1} \frac{\frac{d}{\mathrm{dx}} \tan \left(\frac{\pi x}{2}\right)}{\frac{d}{\mathrm{dx}} \left(\frac{1}{1 - x}\right)}$

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = {\lim}_{x \to 1} \frac{\frac{\pi}{2 {\cos}^{2} \left(\frac{x \pi}{2}\right)}}{\frac{1}{1 - x} ^ 2}$

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = {\lim}_{x \to 1} \frac{\pi {\left(1 - x\right)}^{2}}{2 {\cos}^{2} \left(\frac{x \pi}{2}\right)}$

This is now in the form $\frac{0}{0}$ and we can apply l'Hospital's rule again:

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = {\lim}_{x \to 1} \frac{\frac{d}{\mathrm{dx}} \left(\pi {\left(1 - x\right)}^{2}\right)}{\frac{d}{\mathrm{dx}} \left(2 {\cos}^{2} \left(\frac{x \pi}{2}\right)\right)}$

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = {\lim}_{x \to 1} \frac{- 2 \pi \left(1 - x\right)}{- 2 \pi \cos \left(\frac{x \pi}{2}\right) \sin \left(\frac{x \pi}{2}\right)} = 2 {\lim}_{x \to 1} \frac{1 - x}{\sin} \left(\pi x\right)$

and again:

${\lim}_{x \to 1} \tan \frac{\frac{\pi x}{2}}{\frac{1}{1 - x}} = 2 {\lim}_{x \to 1} \frac{\frac{d}{\mathrm{dx}} \left(1 - x\right)}{\frac{d}{\mathrm{dx}} \sin \left(\pi x\right)} = 2 {\lim}_{x \to 1} - \frac{1}{\pi \cos \pi x} = \frac{2}{\pi}$